[Quintus tractatus]
◉Liber iste in duas partes partitus est. Prima pars est prohemium libri; secunda in ymaginibus. 
◉This book is divided into two parts. The first part constitutes the book’s prologue; the second [focuses] on images. 
◉[Capitulum 1] 
◉[Chapter 1] 
Prima pars 

◉ Liquet ex libro quarto quod forme rerum visarum reflectuntur ex corporibus politis, et visus adquirit eas in corporibus politis propter reflexionem. Et patuit quomodo fiat adquisitio rerum ex reflexione formarum. Et visus comprehendit rem visam in loco reflexionis determinato et primo cum non fuerit situs rei vise ad visum mutatio. Et forma in corpore polito comprehensa nominatur ymago. Et nos explanabimus in hoc libro loca ymaginum ex corporibus politis, et dicemus quomodo adquiratur horum locorum scientia, et quomodo inveniantur sillogistice, et demonstratur. 
◉ It is clear from the fourth book that the forms of visible objects are reflected from polished bodies, and sight apprehends them in polished bodies according to reflection. And it has been shown how objects are apprehended through the reflection of [their] forms. Moreover, sight perceives the visible object at a determinate and principal location of reflection when there is no change in the spatial disposition of the visible object with respect to the eye. The form perceived in a polished body is called an image. And in this book we shall explain imagelocations in polished bodies, and we shall discuss how to gain a scientific understanding of these locations, as well as how to establish that science by demonstration, and [we shall also discuss] how [those] imagelocations are rationally determined. 
◉[Capitulum 2] 
◉[Chapter 2] 
Pars secunda: loqui in ymaginibus. 

◉ Ymaginis cuiuscumque puncti locus est punctus in quo linea reflexionis secat perpendicularem a puncto rei vise intellectam super lineam contingentem lineam communem superficiei speculi et superficiei reflexionis aut superficiei speculo continue et superficiei reflexionis. Et nos declarabimus. 
◉ The imagelocation of any point is the point where the line of reflection intersects the normal imagined [to extend] from a point on a visible object to the line tangent to the common section of the surface of the mirror and the plane of reflection, or [to the common section] of the plane that coincides with [the plane of the] mirror and the plane of reflection. We shall demonstrate [this point as follows].⁑ 
◉ Sumatur speculum planum, et statuatur equidistans orizonti, et lignum directum et politum ortogonaliter erigatur supra speculum. Et sit speculi quantitas ut totum possit videri lignum, nisi enim totum apparuerit, error inerit. Et signetur in ligno punctum aliquod nigrum. Apparebit quidem visui lignum huic equale ultra speculum huic ligno continuum et ortogonale supra speculum, et in ligno apparenti apparebit punctus signatus tantum distans a superficie speculi quantum ab eadem distat in ligno superiori. Et si declinetur lignum supra speculum, apparebit apparens eadem declinatione declinatum, et punctus signatus in apparenti signato eque remotus a superficie speculi. Et si a puncto signato lignum aliquod erigatur ortogonaliter supra speculum, videbitur hoc etiam lignum a puncto apparenti ortogonaliter supra speculum et huic ortogonali continuum. Idem accidit pluribus punctis in ligno signatis. Idem penitus accidet elevato aut depresso speculo. 
◉ Take a plane mirror, set it up horizontally, and stand a smooth, straight wooden rod upright upon the mirror. Let the mirror be large enough that the whole rod can be seen [in it], for, unless the whole rod is visible, error can arise. Let a black spot be marked on the rod. To the eye a rod the same size as this one will appear behind the mirror and directly in line with the actual rod and orthogonal to the mirror, and in the rod that is seen [in the mirror], the black spot marked [on the actual rod] will appear to lie the same distance behind the mirror’s surface as [it lies] on the rod above that surface. Furthermore, if the [actual] rod is inclined to the mirror, the rod seen [in the mirror] will appear inclined with the same slant, and the spot marked on the rod seen [in the mirror] will appear to lie the same distance from the mirror’s surface as the spot marked [on the actual rod]. Moreover, if some [other] rod is dropped orthogonally to the mirror from the spot marked [on the actual rod], this rod will also appear [behind and] orthogonal to the mirror from the point on the apparent rod, and [it will appear] right in line with the actual rod. The same thing happens for multiple points marked on the rod. The very same thing will happen if the mirror is raised or lowered [by tilting]. 
◉ Planum ergo per hoc quod ymago puncti visi apparet in perpendiculari ducta a puncto viso ad superficiem speculi, et in hoc speculo que perpendicularis est super superficiem speculi est perpendicularis super lineam communem superficiei speculi et reflexionis. 
◉ From this, then, it is evident that the image of a visible point appears on the normal extended from the visible point to the surface of the mirror, and that in this mirror what is perpendicular to the mirror’s surface is perpendicular to the common section of the mirror’s surface and the [plane of] reflection. 
◉ Idem patere poterit in piramide super basim ortogonali, cuius basis plana speculo plano ortogonaliter sit adhibita, apparebit enim huic piramis alia continua quarum eadem basis et harum piramidum acumina equaliter a speculo distantia. Et planum quod, si ab acumine ad acumen ducatur linea recta, erit perpendicularis super basim, et ita super speculum, cum eadem sit superficies speculi et basis, quare conus piramidis in perpendiculari videbitur ab eo ad speculum ducta. Similiter a quocumque puncto piramidis ducatur linea ad punctum respiciens ipsum in apparenti piramide. Erit linea ortogonalis super basim et super speculi superficiem, quia ymago cuiuscumque puncti piramidis cadit in perpendiculari intellecta a puncto illo in speculi superficiem. 
◉ The same thing can be shown in the case of a right cone placed upright with its base lying flat upon the plane mirror, for [in that case] another cone will appear [behind the mirror] directly in line with it and sharing the same base, the vertices of these cones lying the same distance from the [surface of the] mirror. And it is obvious that, if a straight line is extended from vertex to vertex, it will be perpendicular to the [shared] base, and thus to the [surface of the] mirror, since that surface and the base [of the cones] coincide, so the [image of] the cone’s vertex will appear on the normal dropped from that vertex to the mirror. Likewise, from some point on the cone draw a line to the corresponding point on the cone seen [behind the mirror]. That line will be orthogonal to both the [cone’s] base and the mirror’s surface, so the image of any point on the cone lies on the perpendicular imagined [to extend] from that point to the mirror’s surface.⁑ 
◉ Sed quicumque corporis punctus opponatur speculo plano est intelligere piramidem cuius punctus ille sit conus, que quidem piramis super basim ortogonalis, et etiam super speculi superficiem aut ei continuam. Et est intelligere aliam huic piramidi oppositam quarum basis eadem et super speculum ortogonalis, et perpendicularis a cono ad conum ortogonalis erit supra speculum, quare ymago cuiusque puncti speculo quidem oppositi cadit in perpendiculari a puncto ad speculi superficiem aut ei continuam. Sed planum quod in speculis non accedet comprehensio formarum nisi per lineas reflexionum, quare ymago puncti visi cadit in lineam reflexionis, et quelibet talis linea est recta, quare ymago cuiuscumque puncti cadit in punctum sectionis perpendicularis ab illo puncto in superficiem speculi et linee reflexionis. Et in speculis planis cadit, et unica est linea communis superficiei speculi et superficiei reflexionis cum linea contingente locum reflexionis, quare planum quod in speculis planis proprius ymaginis est locus punctus sectionis perpendicularis a puncto visi super lineam contingentem communem lineam superficiei speculi et superficiei reflexionis et linee reflexionis. 
◉ But whatever point on a body may face a plane mirror, a cone can be imagined with its vertex at that point, and this cone is orthogonal to its base as well as to the surface of the mirror or its continuation. And another cone opposite this one can be imagined sharing the same base and orthogonal to the mirror, and the perpendicular [extended] from vertex to vertex [of the two cones] will be orthogonal to the mirror, so the image of any point facing the mirror lies on the normal [extended] from the point to the surface of the mirror or to its extension. But it is obvious that in [all] mirrors the perception of forms will be attained only along lines of reflection, so the image of the point seen [in the mirror] lies on a line of reflection, and every such line is straight, so the image of any point [seen in a mirror] lies at the intersectionpoint of the normal [dropped] from that point to the surface of the mirror and the line of reflection. And it falls on plane mirrors, and the common section of the mirror’s surface and the plane of reflection is unique [in coinciding] with the line tangent to the point of reflection, so it is evident that in plane mirrors the appropriate imagelocation is the point where the normal [dropped] from the visible point to the line tangent to the common section of the mirror’s surface and the plane of reflection intersects the line of reflection.⁑ 
◉ In speculis spericis et extra politis patebit quod diximus. Queratur superficies speculi talis magna in qua appareat forma baculi gracilis perpendiculariter erecta super ipsum. Apparebit quidem forma baculi baculo continua, et apparebit in forma baculi punctus signatus distans a superficie speculi secundum distantiam eius ab eodem in baculo. Et si fuerit baculus gracilior ex parte unius capitis quam ex parte alterius, apparebit quidem in hoc speculo forma eius piramidalis, et est error visus quem postea assignabimus. 
◉ In the case of spherical mirrors that are polished on the outer surface [i.e., convex] what we have claimed will be evident. Find [such] a mirror with a surface large enough that the [entire] form of the thin rod [just used with the plane mirror] can be seen when the rod is stood upright on it. The form of the rod [seen in the mirror] will appear directly in line with the rod [itself], and on the form of the rod [the form of] the point marked on it will appear [to lie] the same [relative] distance behind the mirror’s surface as it lies [above the mirror] on the actual rod.⁑ Moreover, if the rod is narrower at one end than at the other, its form will appear conical in this kind of mirror, but this involves a visual illusion that we will explain later.⁑ 
◉ Amplius, fiat piramis cum eo ortogonalis super basim circularem circulatione perfecta, et applicetur etiam huic speculo. Videbitur quidem piramis huic continua super eandem basim erecta, sed minor ista. Quod appareat piramis planum per hoc quod omnes linee ab apparenti ymagine coni ad circulum basis videantur equales, et si declinetur piramis modicum supra speculum a situ in quo tota videtur, ut scilicet aliquid ex eo abscondatur, dum tamen locus reflexionis in speculo visui exponatur, apparebit inde ymago piramidis. Et si elongetur visus a speculo aut accedat dum tamen super lineam a loco reflexionis ad ipsum protractatum cedat, comprehendetur ymago piramidis, sed et accessus vel recessus secundum hanc lineam erit ut notetur locus reflexionis. Et a nota ad locum visus ducatur linea secundum quam processus fiat. 
◉ Now, along with this rod form a cone that is orthogonal to a circular base that is rounded off perfectly, and let it also be applied to this mirror. A cone continuous with this one will appear [behind the mirror and] standing on the same base, but [it will appear] smaller than the [actual] cone. That the image should appear conical is clear from the fact that all the lines [extended] from the vertex of the cone seen [in the mirror] to the circle at the base appear equal, and if the cone is moved a bit on the mirror’s surface away from where all of it is visible, i.e., so that some of it disappears from view, then, as long as the point of reflection on the mirror is exposed to sight, the image of the cone will be visible.⁑ And if the eye is drawn away from the mirror or approaches it, provided it follows the [original] line of reflection extending between it and the point of reflection, the image of the cone will be visible, but the movement [of the eye] toward or away will be along this line in such a way that the point of reflection is marked. Extend a line from the location of the eye to the mark according to which the movement occurs. 
◉ Verum quoniam ymago piramidis ortogonalis super basim piramidis, et basis est circulus ex circulis in spera, erit linea a cono piramidis ad conum ymaginis ducta ortogonalis super circulum illum, et transibit per centrum eius. Et erit ortogonalis super speram et transibit per centrum spere, et erit ortogonalis super superficiem speram contingentem in puncto per quem transit hec linea. Et erit similiter ortogonalis super lineam contingentem circulum spere per punctum illum transeuntem, et hec contingens est linea communis superficiei reflexionis et superficiei contingentis speram in puncto illo, et hec linea est contingens circulo spere communi superficiei spere et superficiei reflexionis. Linea ergo a cono piramidis ad conum ymaginis ducta est perpendicularis super lineam contingentem lineam communem superficiei reflexionis et superficiei speculi, que quidem est circulus. 
◉ Now, since the image of the cone is orthogonal to the cone’s base, and since that base forms one of [an infinite number of possible] circles on the sphere, the line [extended] from the vertex of the [actual] cone to the vertex of its image will be orthogonal to that circle, and it will pass through its center. It will also be orthogonal to the sphere and will pass through the sphere’s center, and it will be orthogonal to the plane tangent to the sphere at the point where this line passes through [its surface]. Likewise, it will be orthogonal to the line tangent to the [great] circle on the sphere that passes through that point, and this tangent forms the common section of the plane of reflection and the plane tangent to the sphere at that point, and this line is tangent to the circle on the sphere that forms the common section of the sphere’s surface and the plane of reflection. Thus, the line extended from the vertex of the cone to the vertex of its image is perpendicular to the line tangent to the common section of the plane of reflection and the mirror’s surface, and that section is a circle. 
◉ In hac igitur perpendiculari videtur ymago coni, et planum quod ymago coni est in linea reflexionis, quare comprehendetur ymago coni in concursu linee reflexionis et perpendicularis a cono ad speram ducte sive ad contingentem circulum communem superficiei spere et reflexionis. Sumpto autem quocumque puncto huic speculo opposito est intelligere piramidem super superficiem speculi ortogonalem aut super continuam, cuius conus sit punctus sumptus. Et linea ab illo puncto ad ymaginem puncti illius erit in superficie reflexionis et perpendicularis super superficiem speculi vel ei continuam modo predicto, quoniam punctus visus et ymago eius semper sunt in superficie reflexionis, quare et linea a puncto viso ad eius ymaginem ducta. 
◉ Therefore, the image of the vertex will be seen on this perpendicular, and it is evident that the image of the vertex lies on the line of reflection, so the image of the vertex will be perceived at the intersection of the line of reflection and the normal dropped from the [actual] vertex to the sphere or to the tangent to the circle forming the common section of the sphere’s surface and the plane of reflection. Furthermore, from any given point facing this [sort of] mirror a cone can be imagined orthogonal to the mirror’s surface or to its extension, and the vertex of this cone is the given point. The line from that point to the image of that point will lie in the plane of reflection as well as on the normal to the mirror’s surface or its extension in the way discussed before, for the point seen and its image always lie in the plane of reflection, and so does the line extended from the point seen to its image. 
◉ In speculis columpnaribus exterius non apparent que in ligno et piramide diximus, quoniam recta in hiis speculis videtur non recta, et est error visus cuius postea causam assignabimus. Accidit tamen in solo corporis puncto videre locum ymaginis predictum. 
◉ In the case of cylindrical mirrors [polished] on the outer surface, the things we claimed about the rod and the cone [in the previous example] are not observed, because in these mirrors what is straight does not appear straight, and there occurs a visual error whose cause we will explain later.⁑ Still, for [any] single point on a visible object [seen in such a mirror] the imagelocation that has been described can be observed. 
◉ Verbi gratia, adhibito precedentis libri instrumento, immittatur regula cui sit infixum columpnare speculum ut media portionis speculi linea sit in superficie regule. Et non transeat hec regula tabulam eneam sed super ipsum cadat ortogonaliter ita quod altitudo regule sit super lineam dividentem triangulum tabule enee. Erectio facta in hac tabula, impleatur cera, et inducatur ei planities ut sit in eadem superficie cum tabula, et est ut certior fiat ortogonalis regule directio super tabulam. 
◉ For instance, having set up the apparatus [described] in the previous book, insert the panel with the [convex] cylindrical mirror attached in it so that the midline of the portion of the mirror [contained by the panel] lies in the plane of the panel. And do not let this panel pass [below] the bronze plaque, but let it stand orthogonally to it so that the [bottom endpoint of] the line along the height of the panel stands on the line that bisects the triangle of the bronze plaque. When the panel is set up on this plaque, fill [the space below] with wax and level the wax so that it lies in the same plane as the [top surface of the] plaque, and this is [done] so that the panel can be placed perfectly perpendicular to the plaque.⁑ 
◉ Deinde queratur regula acuta, et acuatur extremitas, et applicetur huius regule acuitas medie superficiei anuli linee. Et descendat super hanc lineam, et ubi ceciderit super regulam fiat signum. Postea acus descendat super hanc lineam in qua sit infixum modicum corpus album, et hoc in termino, nec descendat acus usque ad regulam. 
◉ Then find a pointed ruler, sharpen its edge, and lay this sharpened edge along the midline of the surface of the [cylindrical] ring.⁑ Let it reach along this line [to the mirror], and mark the spot where it touches the [mirror in the] panel. Then lay a needle along this line, and on its end let a small white object be attached, and do not let the needle reach all the way to the [mirror in the] panel. 
◉ Adhibeatur autem visus ut sit in superficie regule, et claudatur unus visuum. Videbitur quidem ymago corporis super lineam a puncto signato ad acumen acus protractam, que quidem linea perpendicularis est super superficiem regule, que superficies tangit columpnam in linea longitudinis; et est perpendicularis super lineam longitudinis columpne, que est in superficie regule et est linea communis superficiei columpne et superficiei reflexionis. Et in superficie reflexionis sunt linea longitudinis et linea perpendicularis. 
◉ Let the eye[s] be placed in the plane of the ruler, and let one of them be closed. The image of the [small white] object [at the end of the needle] will be seen along the line extended from the point marked [on the mirror] to the point of the needle, and that line is of course perpendicular to the surface of the panel, which is tangent to the cylinder [from which the mirror is formed] along its line of longitude; and that line is perpendicular to the cylinder’s line of longitude, which lies in the plane of the panel and forms the common section of the cylinder’s surface and the plane of reflection. Also the line of [the cylinder’s] longitude and the line perpendicular [to it] lie in the plane of reflection. 
◉ Et si situs visus mutetur, et circa anuli superficiem visus volvatur, apparebunt sicut prius, et in eadem linea corpus et ymago corporis et acus. Et est linea illa perpendicularis super mediam longitudinis columpne lineam, et est hec perpendicularis in superficie reflexionis, quoniam superficies anuli secat columpnam super circulum equidistantem basi columpne, et in hac superficie est visus. Et nos probabimus postea quod, quando visus et visum corpus fuerint in superficie equidistanti basi columpne, illa est superficies reflexionis. In hoc autem situ linea communis superficiei columpne et superficiei reflexionis est circulus, et perpendicularis in qua videntur ymago et corpus ortogonaliter cadit super lineam super circulum contingentem. 
◉ Now, if the location of the eye is shifted, and if it moves along [the outer edge of the upper] surface of the [cylindrical] ring, then, just as before, the [small white] object [at the end of the needle], the image of the object, and the [point of the] needle will appear [to lie] on the same line. Moreover, that line is perpendicular to the midline along the length of the cylinder, and this perpendicular lies in the plane of reflection, for the [top] surface of the [cylindrical] ring intersects the cylinder along a circle that is parallel to the base of the cylinder, and the eye lies in the plane [of this circle]. And later on we shall demonstrate that, when the eye and the visible object lie in a plane parallel to the base of the cylinder, that plane constitutes the plane of reflection.⁑ In this situation, however, the common section of the cylinder’s surface and the plane of reflection is a circle, and the normal upon which the image and the [small white] object are seen falls orthogonally to a line tangent to the circle.⁑ 
◉ Hiis peractis, auferatur acus a loco suo, et ponatur regula acuta super lineam mediam ita quod cadat super mediam longitudinis regule lineam, et adhibeatur regula acuta superficiei anuli cera firmiter. Postea auferatur regula in qua est speculum, et accipiatur regula acuta, et applicetur eius acuitas medie longitudinis regule linee, et secundum processum acuitatis fiat cum incausto super speculum protractio. Post sumatur triangulus cereus modicus cuius unum latus sit equale altitudini regule in qua est speculum, et sit altitudo huius trianguli moderata, et superficies huius trianguli sit plane pro posse. Et adhibeatur columpne regule tabule ceree triangulus firmiter sub base regule, et latus eius equale altitudini regule ponatur super latus basis regule. Cum ita fuerit, erit huius trianguli altitudo super basem columpnalis equalem tabule regule, et ut efficiatur superficies plana ad modum superficiei regule, includatur triangulus inter regulam et superficiem planam, et comprimatur donec sit bene planitus. Et super superficiem huius trianguli ponatur regula acuta, et secetur finis huius trianguli cum acuitate regule, et erit finis eius linea recta. Et erit linea hec basis regule in qua est speculum. 
◉ When all this is done, remove the needle and place the sharpedged ruler along the midline [of the cylindrical ring’s upper surface] so that it reaches the midline along the longitude of the panel, and attach the sharpedged ruler firmly to the [cylindrical] ring’s [upper] surface with wax. Then remove the panel with the mirror in it and take a pointed ruler and lay its sharpened edge upon the midline along the length of the panel, and along the sharpened edge draw a line upon the mirror with ink. Then take a moderatesized wax triangle, one of whose sides is equal to the height of the panel containing the mirror, let it be moderately thick, and let the surface of this triangle be as flat as possible. Then attach the wax triangle firmly to the panel containing the cylinder below the panel’s base, and place the side of the triangle that is equal to the height of the panel upon the side of the panel’s base. When that is done, the length of this triangle at the base of the [panel containing the] cylinder will be equal to the [height of the] panel, and in order to make the triangle’s surface as flat as the surface of the panel, place the triangle between a panel and a flat surface and compress it until it is quite flat. Place a sharpedged ruler on the surface of this triangle, and cut off the bottom edge of this triangle along the edge of the ruler, and this edge will form a straight line. This line will form the base for the panel containing the mirror. 
◉ Postea ponatur regula super superficiem tabule que est in instrumento, et ponatur finis basis eius, que est in longitudine que est latus trianguli cerei, super lineam que est in longitudine eris, sicut factum est prius. Et erit superficies regule in qua est speculum ortogonalis super tabulam eneam, et hec superficies secat tabulam eneam super lineam que est in longitudine eris, et hec superficies tangit superficiem speculi super lineam que est in superficie speculi. Et hec superficies est superficies regule in qua est speculum, et erit angulus regule acute adherentis in media linea superficiei anuli in qua superficie erit speculum declinatum in partem in qua est caput trianguli, quia regula exaltavit unam partem eius cum corpore trianguli, et alia pars que est post caput trianguli est superficies eris, et erit linea que est in medietate speculi declinata. 
◉ Then place the panel on the surface of the [bronze] plaque in the instrument, and place the edge of its base, which lies along the length formed by the side of the wax triangle, on the line of longitude of the bronze [plaque], as was done before. The plane of the panel containing the mirror will be perpendicular to the bronze plaque, and this plane cuts the bronze plaque along the line of longitude of the bronze plaque, and this plane is tangent to the surface of the mirror along the line on the mirror’s surface. And this plane is the surface of the panel containing the mirror, and the angle [along the edge] of the pointed ruler will be attached along the midline of the [top] surface of the ring, and the mirror’s surface will be tilted downward to that surface on the side of the triangle’s vertex, since one side of the panel has been raised according to the breadth of the triangle, while the other side [of the triangle] beyond the vertex of the triangle is the plane of the bronze [plaque], and the line that lies along the midline of the mirror will be inclined. 
◉ Et quando fuerit latus trianguli cerei super lineam que est in longitudine eris, movebitur regula in qua est speculum, et latus trianguli in hoc motu, si sit super lineam longitudinis eris. Et procedat vel retrocedat donec concurrat angulus regule acute cum puncto aliquo linee superficiei speculi donec firmetur regula acuta, et auferatur linea in speculo cum incausto facta. Et fiat punctus in superficie speculi in directo capitis regule acute. Et auferatur regula acuta, et apponatur acus, et sit acus super lineam mediam superficiei anuli, et adherere cogatur cum cera. Et erit linea intellectualis ab acu in punctum signatum in superficie speculi perpendicularis super superficiem regule que tangit superficiem speculi super punctum signatum et perpendiculariter super quamlibet lineam ab illo puncto protractam in superficiem contingentem speculum. Erit igitur perpendicularis super lineam rectam contingentem lineam communem superficiei alte anuli et superficiei. 
◉ And when the side of the wax triangle is [positioned] along the line of longitude of the bronze [plaque], the panel that contains the mirror will move along the line of longitude of the bronze plaque, and the [base] side of the [wax] triangle will move with it, if it is [positioned] on the [bronze plaque’s] line of longitude. And let it move back and forth until the point of the sharpedged ruler meets some point on the [mid]line on the surface of the mirror and the mirror fits snugly against the sharpedged ruler, and erase the line drawn in ink on the mirror. Mark the point on the surface of the mirror that is directly under the end of the sharpedged ruler. And remove the sharpedged ruler, and apply the needle, and let the needle lie along the midline of the [top] surface of the ring, and let it be attached firmly with wax. And the imaginary line from the point of the needle to the point marked on the surface of the mirror will be perpendicular to the plane of the panel which is tangent to the mirror’s surface at the marked point, and it will lie perpendicular to any line extended through that point in the plane tangent to the mirror. Thus, it will be perpendicular to the straight line tangent to the common section of the top surface of the ring and the [mirror’s] surface. 
◉ Ponatur autem visus in superficie anuli in capite eius et videbit in speculo donec comprehendat formam corporis parvi quod est in acu, et tunc percipiet corpus illud, et punctum in speculo signatum, et ymaginem illius corporis. Et linea transiens per corpus parvum et per punctum signatum in superficie est perpendicularis super superficiem contingentem speculi superficiem super punctum signatum. Et hec superficies anuli est ex superficiebus reflexionis, et corpus parvum et centra visus sunt in hac superficie, et punctus reflexionis est in hac superficie, et hoc deinceps probabimus. Et ymago corporis parvi in hoc situ erit super lineam rectam a corpore parvo protractam rectam super superficiem contingentem superficiem speculi, et cum hec linea perpendicularis super lineam rectam contingentem lineam communem superficiei speculi et superficiei reflexionis que est superficies anuli, et superficies reflexionis est ex superficiebus declinantibus secantibus columpnam inter lineas longitudinis columpne et circulos eius equidistantes basibus, quia regula et speculum quod est in ea sunt declinata, linea ergo communis huic superficiei et superficiei speculi est ex sectionibus columpnaribus. Et ita explanabimus locum ymaginis si mutetur situs regule in qua est speculum et declinetur super superficiem eius alia declinatione minori vel maiori. 
◉ Now, place the eye in the plane of the ring at its apex [on the midline along its top surface], and it will look into the mirror until it perceives the form of the small [white] object on the needle, and then it will perceive that object, the point marked on the mirror, and the image of that object [all in a line]. And the line passing through the small [white] object and the point marked on the [mirror’s] surface is perpendicular to the plane tangent to the mirror’s surface at the point marked on it. And this surface on the ring is among the planes of reflection, and the small [white] object and the centers of sight lie in that plane, and the point of reflection is in that plane, and we will prove this later on. And in this case the image of the small [white] object will lie upon a straight line extended from the small object [itself] directly to the plane tangent to the mirror’s surface, and since this line is perpendicular to the straight line tangent to the common section of the surface of the mirror and the plane of reflection, which is the [top] surface of the ring, and since the plane of reflection is among the oblique planes that cut the cylinder between the lines of longitude on the cylinder and the circles parallel to its bases, then, because the panel and the mirror in it are at a slant, the common section of this plane [of reflection] and the surface of the mirror is among the cylindrical sections [i.e., an ellipse].⁑ And we will explain the imagelocation in this same way if the orientation of the panel containing the mirror is changed and slanted on its surface at a smaller or greater obliquity. 
◉ Palam ergo ex hiis quod ymago percipitur ubi perpendicularis a viso puncto ad speculi superficiem ducta concurrit cum linea reflexionis, et hic est situs predictus. Si a puncto viso ad speculi superficiem ducantur linee ad speculi superficiem, que perpendicularis est minor qualibet alia, quoniam quelibet alia prius secat lineam communem superficiei contingenti speculum in qua ortogonaliter cadit perpendicularis et huic superficiei reflexionis quam veniat ad speculum, et quelibet linea a puncto viso in hac superficie ad hanc lineam communem ducta est maior perpendiculari, quia maiorem respicit angulum, quare propositum. 
◉ It is therefore evident from these things that the image is perceived where the normal extended from the visible point to the surface of the mirror intersects the line of reflection, and this is the predicted location. If lines are drawn from the visible point to the mirror’s surface, the one that is perpendicular is shorter than any of the others, for any of the others first intersects the common section of the plane tangent to the mirror’s surface to which the perpendicular falls and the plane of reflection that reaches the mirror, and any line extending in this plane from the visible point to this common section is longer than the perpendicular, since it subtends a greater angle, so what was proposed [is demonstrated].⁑ 
◉ Eadem poterit adhibi comparatio in speculo piramidali exteriori, et idem patebit sive sint ymagines rerum visarum in sectionibus piramidalibus, sive in eis que fuerint secundum lineas longitudinis. 
◉ The same procedure can be applied in the case of the conical mirror [polished] on the outer surface, and the same thing will be shown whether the images of objects are seen in [planes formed by] conic sections, or whether they are seen in [planes formed by] cuts made along lines of longitude.⁑ 
◉ In speculis spericis concavis comprehenduntur ymagines quedam ultra speculum, quedam in superficie, quedam citra superficiem, et harum quedam comprehenduntur in veritate, quedam preter veritatem. 
◉ In concave spherical mirrors some images are perceived behind the mirror, some on the mirror’s surface, and some in front of the mirror, and some of these images are perceived according to reality, whereas some are perceived in ways that do not accord with reality.⁑ 
◉ Omnes quarum comprehenditur veritas apparent in loco sectionis perpendicularis et linee reflexionis, quod sic patebit. Fiat piramis, et ea ortogonalis super basem, et dyameter basis sit minor medietate dyametri spere, et linea longitudinis piramidis sit maior eadem semidyametro. Et secetur ex parte basis ad quantitatem eius, et fiat super sectionem circulus, et secetur piramis super hunc circulum. Postea in medio speculi fiat circulus ad quantitatem basis piramidis remanentis, et aptetur huic circulo piramis, et firmetur cum cera. 
◉ That all of those images seen according to reality appear at the point where the normal [dropped from the object] intersects the line of reflection will be demonstrated as follows. Fashion a cone that is orthogonal to its base [i.e., a right cone], and let the diameter of the base be smaller than the radius of the sphere [from which the concave mirror is formed], and let the line along the length of the cone be longer than that radius. Now, [from the vertex] toward the base, cut off a length [along the cone’s line of longitude] equal to that radius, inscribe a circular section [parallel to the cone’s base at that point] and cut the cone according to this circle. Then, in the middle of the mirror draw a circle the same size as the base of the cone that is left [after the cut], fit the cone to this circle, and attach it firmly with wax. 
◉ Deinde statuatur visus in situ in quo ymaginem piramidis possit comprehendere, et adhibeatur lux ut certior fiat comprehensio. Non videbis quidem piramidem huic coniunctam, sed comprehendes hanc ultra speculum extensam, unde apparebit piramis quedam continua cuius basis ultra speculum et pars eius piramis cerea. Et si in hac piramide signetur linea longitudinis cum incausto, videbitur hec linea protendi super superficiem piramidis apparentis, et quoniam conus piramidis est centrum spere, linea a cono secundum longitudinem piramidis ducta erit perpendicularis super contingentem cuiuslibet circuli spere per caput linee transeuntis. 
◉ Place the center of sight where the image of the cone can be perceived, and direct light upon it so that the perception may be clearer.⁑ You will not see a cone conjoined with this one, but you will perceive [the image of] this [cone] extending behind the mirror, so a certain cone directly in line [with it] will appear with its base behind the mirror along with the wax section of its cone.⁑ And if a line of longitude is marked with ink on this cone, this line will appear to extend along the surface of the cone that appears [in the mirror], and since the vertex of the cone lies at the center of the sphere, the line extending from the vertex along the length of the cone will be perpendicular to the tangent to any circle on the sphere that passes through the endpoint of the line. 
◉ Quare quelibet linea longitudinis piramidis apparentis est perpendicularis super lineam contingentem lineam communem superficiei reflexionis et superficiei spere, que quidem linea est communis, et est circulus. Et quilibet punctus piramidis in hac videtur perpendiculari, et quelibet perpendicularis est in superficie reflexionis, quoniam punctus visus et ymago eius sunt in perpendiculari et in hac superficie reflexionis. Et omnis ymago comprehenditur in linea reflexionis, quare ymago cuiuscumque puncti piramidis erit in puncto sectionis perpendicularis et linee reflexionis. 
◉ Hence, any line of longitude on the cone that appears [in the mirror] is perpendicular to the line tangent to the common section of the plane of reflection and the surface of the sphere, that line being the common section, and it is a [great] circle. And any point on the cone [lying] on this perpendicular is seen, and every [such] perpendicular lies in the plane of reflection, because the visible point and its image lie on the normal as well as in this plane of reflection. But every image is perceived along the line of reflection, so the image of any point on the cone will lie at the point of intersection of the normal and the line of reflection. 
◉ Puncta autem quorum ymagines citra speculum comprehenduntur, hoc est inter visum et speculum sunt, cum a quolibet eorum linea ducta ad centrum speculi secet latitudinem vie visum et speculum interiacentis. Et ut videatur hoc, auferatur piramis a medio speculi, et collocetur in parte. Erit conus centrum speculi, et remotio visus a speculi superficie sit maior semidyametro spere. Deinde sumatur lignum gracile album, et statuatur in speculo ut sit centrum speculi directe medium inter caput ligni et centrum visus, et dirigatur intuitus in punctum speculi a quo linea ad conum piramidis ducta sit inter caput ligni et visum. Et inspiciatur speculum donec non appareat caput ligni et lignum, et apparebit forma capitis ligni citra speculum et propinquior visui cono piramidis. Et erunt in eadem linea recta conus piramidis, et caput ligni, et ymago capitis, et hec linea est perpendicularis super lineam contingentem lineam communem superficiei speculi et superficiei reflexionis. Quoniam superficies reflexionis transit per centrum et punctum visus, et linea transiens per hec duo puncta est in superficie reflexionis, et linea communis est circulus. Et hec linea huic circulo erit dyametrum, quoniam centrum illius circuli est centrum spere, quare erit hec linea perpendicularis super lineam contingentem circulum in capite huius linee, et hec linea transit per punctum visum et per eius ymaginem. Et ita quodlibet punctum citra speculum visum comprehenditur in eadem linea cum centro et cum ymagine eius, et quodlibet punctum videtur in linea reflexionis, quare in loco sectionis perpendicularis et linee reflexionis. 
◉ Furthemore, points whose images are perceived in front of the mirror—i.e., ones that lie between the center of sight and the mirror—[are perceived this way] since the line extended from any of them to the center of the mirror will cut the side of the line lying between the visible object and the mirror. In order to observe this [fact], remove the cone from the center of the mirror and replace it at the side. The vertex will be the center of the mirror, and let the distance between the center of sight and the mirror be greater than the sphere’s radius. Then take a thin white rod, and stand it on the mirror so that the center of the mirror lies directly between the top of the rod and the center of sight, and direct your line of sight to the point on the mirror from which the line extended to the vertex of the cone lies between the top of the rod and the center of sight. Then, look into the mirror until neither the top of the rod nor the rod itself appears, and the form of the top of the rod will appear above the mirror and nearer to the eye than the vertex of the cone. And the vertex of the cone, the top of the rod, and the image of the top of the rod will lie on the same straight line, and this line is perpendicular to the line tangent to the common section of the mirror’s surface and the plane of reflection. For the plane of reflection passes through the center [of sight] and the visible point, and the line passing through these two points lies in the plane of reflection, and the common section is a circle. And this line will form a diameter in this circle, because the center of that circle is the center of the sphere, so this line will be perpendicular to the line tangent to the circle at the endpoint of this line, and this line passes through the visible point as well as through its image. And so, any point [whose image is seen] in front of the mirror is perceived on the same line as the center [of the sphere] and the image, and any point is seen on the line of reflection, so [it is seen] at the point of intersection of the normal and the line of reflection. 
◉ Et ea quorum veritas comprehenditur in hiis speculis sunt quorum ymagines apparent ultra speculum vel citra superficiem eius, et preter hec nulla sunt que in hoc speculo in veritate comprehendat visus, ipsa enim prohibent ymagines suas veras apparere. Ymagines que apparent in superficie speculi huius sunt ex ultima partitione, et hoc explanabimus, cum aderit sermo in erroribus visus. Quodlibet ergo punctum in veritate in hoc speculo comprehensum apparet in concursu perpendicularis et linee reflexionis, que quidem perpendicularis transit a puncto viso ad centrum spere et cadit ortogonaliter in contingentem lineam communem. 
◉ And the things that are perceived in these mirrors as they exist in reality are those whose images appear behind the mirror or in front of the mirror, and beyond these [types of images] there are none that sight may perceive as they exist in reality in this [sort of] mirror, for these latter types of objects do not permit their true images to appear. Images that appear on the surface of this [type of] mirror belong to the last category, and we will explain this when the discussion turns to visual errors. Therefore, any point that is perceived as it actually exists in this [sort of] mirror appears at the intersection of the normal and the line of reflection, and this normal passes from the visible point to the center of the sphere and falls orthogonally on the tangent to the common section [of the mirror and the plane of reflection].⁑ 
◉ In speculis columpnaribus concavis diversificatur ymago, aliquando enim erit locus eius in superficie speculi, aliquando ultra, aliquando citra. Et in hiis omnibus aliquando in veritate comprehenditur, aliquando non. 
◉ In the case of concave cylindrical mirrors the image varies, for its location will sometimes be on the surface of the mirror, sometimes behind it, and sometimes in front of it. And in all these cases [the image] is sometimes perceived as it actually exists, sometimes not. 
◉ Cum volueris in hiis locum ymaginis percipere, facias sicut fecisti in columpnis exterioribus. Adhibeatur enim regula in qua sit columpna concava sicut adhibita est superius, et acus similiter, et corpus modicum in summitate acus. Et ponatur visus oppositus in medio circuli et in medio superficiei anuli, et sublevetur visus modicum a superficie anuli, et inspiciat donec ymaginem corporis videat et comprehendat formam corporis, et corpus, et punctum in speculo signatum in eadem linea perpendiculari super superficiem speculi—et hoc per sillogismum sensualem. Et erit ymago ultra speculum, et erit reflexio ex puncto linee que est in medio speculi. 
◉ If you want to observe the imagelocation in these cases, do what you did [before] in the case of cylindrical mirrors [polished on the] outer surface. Insert the panel with the concave cylindrical mirror on it into the ring, as it was inserted earlier, and likewise [position] the needle and the small [white] object at the end of the needle. Place your eye directly in front of the center of the circle [passing through the middle of the mirror] and at the middle of the [top] surface of the ring, raise the eye a little bit above the ring’s [top] surface, and let it look until it sees the image of the object and perceives the form of the object, the object itself, and the point marked on the mirror along the same line that is perpendicular to the mirror’s surface—and [determine] this according to senseinduction.⁑ The image will lie behind the mirror, and the reflection will occur from a point on the line [of longitude] that lies in the middle of the mirror. 
◉ Deinde statuatur visus in superficie anuli, sed extra medium, donec videat ymaginem corporis parvi. Videbis quidem eam citra speculum, et videbis corpus, et eius ymaginem, et punctum in speculo signatum in una linea recta perpendiculari super lineam rectam contingentem circulum equidistantem basi speculi super punctum signatum in superficie speculi. Et superficies huius circuli est superficies reflexionis in hoc situ, et est superficies faciei anuli, et punctus reflexionis est punctus illius circuli. 
◉ Then place the center of sight in the plane of the [top surface of the] ring, but outside the midline, until it sees the image of the small [white] object [on the needle]. You will see it in front of the mirror, and you will see the object, its image, and the point marked on the mirror along one straight line perpendicular to the straight line tangent to the circle that is parallel to the base of the mirror and [lying] upon the point marked on the mirror’s surface. And the plane of this circle forms the plane of reflection in this case, and it is the top surface of the ring, and the point of reflection is a point on that circle.⁑ 
◉ Postea adhibeatur cum manu alia acus in cuius summitate sit corpus modicum, et statuatur in superficiem et axem hoc modo ut hoc corpus et punctus signatus sint in eadem linea secundum sensualem sillogismum. Et sit visus in superficie anuli inter caput eius et medium. Videbit quidem ymaginem corporis, et videbit hanc ymaginem, et corpus eius, et punctus signatum in superficie speculi in eadem linea recta. 
◉ Afterwards, with your other hand place another needle with a small object attached to its end, and lay it on the [top] surface [of the ring] and the axis in such a way that the object and the point marked [on the mirror] lie on the same line according to senseinduction. Let the eye lie in the plane of the [top surface of the] ring between its endpoint and the middle [of the ring’s top surface]. It will see the image of the object, and it will see this image as well as its [generating] object along with the point marked on the surface of the mirror on the same straight line.⁑ 
◉ Si autem declinetur linea recta cum triangulo parvo quem fecimus—et visus sit in medio anuli—videbis ymaginem citra speculum, sed in eadem linea recta cum corpore et puncto signato. Et hec reflexio erit ex sectionibus columpnaribus, quoniam speculum est declinatum, et scimus quod non percipitur ymago nisi in linea reflexionis. Palam ergo quod locus ymaginis est ubi secat perpendicularis predictam lineam reflexionis, cum comprehenditur veritas, et licet non comprehendatur certitudo ymaginis, tamen erit modus harum ymaginum cum veritatis ymaginibus. 
◉ If, however, the straight line [of longitude in the middle of the panel] is slanted according to the small triangle we fashioned—and the center of sight should be on the midline of the ring—you will see the image in front of the mirror, but on the same straight line as the body and the point marked [on the mirror’s surface]. And this reflection will occur from one of the cylindric sections [i.e., ellipses], because the mirror is inclined, and we know that an image is perceived only along a line of reflection. Therefore, it is clear that the imagelocation lies where the normal intersects the aforementioned line of reflection, since the [image] is perceived properly, and even though the image may not be perceived clearly, these images will still belong to the category of those that are properly seen. 
◉ Pari modo videre poteris in piramidibus concavis in concursu perpendicularis cum lineis reflexionis. Palam ergo quod in omnibus speculis comprehenduntur ymagines in loco predicto, qui quidem locus similiter dicitur ymago. 
◉ In the same way you can observe [the image] in a concave conical mirror at the intersection of the normal with the lines of reflection. It is therefore evident that in all mirrors images are perceived at the aforementioned location, which is likewise referred to as the »image.« 
◉ Quare autem comprehendantur res vise per reflexionem in locis ymaginis et quare ymago sit super perpendicularem a re visa in speculi superficiem declarabimus causam. Visus, cum adquirit formam per reflexionem, adquirit eam statim sine certitudine, et adquirit longitudinem per estimationem. Et hanc longitudinem comprehendet forsitan in veritate per diligentiam intuitus adhibitam, forsitan non. Et istud explanavimus in libro secundo, et ibi dictum est quod visus adquirit longitudinem per sillogismum ex magnitudine corporis et angulo aliquo sub quo comprehenditur magnitudo. Et adquisitio rei site ignote manifeste est in hunc modum. Res etiam note comprehenduntur in hunc modum, conferuntur enim rebus cognitis et magnitudinibus vel longitudinibus notis. Cum visus comprehendit rem aliquam per reflexionem, non comprehendit longitudinem ymaginis nisi per estimationem; deinde adhibita diligentia, adquirit longitudinem, et verificat per sillogismum ex magnitudine rei vise et angulo piramidis super quam forma reflectitur ad visum. 
◉ We will now explain why visible objects are perceived through reflection where the image is located and why the image lies on the normal [dropped] from the visible object to the surface of the mirror. When the visual faculty grasps a form by reflection, it grasps it immediately without an accurate determination, and it grasps its distance through estimation. It may perceive this distance accurately through a close and careful scrutiny, or it may not. And we have explained this [process] in the second book, where it was claimed that the visual faculty grasps distance through a deduction based on the size of the object and the particular angle under which the size is perceived. The perception of a visible object whose location is unknown clearly occurs in this way. Also, objects [whose locations are] known are perceived in this way, for they are compared to known objects and known sizes or distances.⁑ When sight perceives some object through reflection, it perceives the distance of the image solely by estimation; then, after a close scrutiny, it grasps the distance and verifies [it] deductively on the basis of the size of the visible object and the angle of the cone according to which the form is reflected to the eye. 
◉ Cum ergo res visa ex rebus notis fuerit, visus adquirit eius longitudinem per iam notam longitudinem equalem angulum huic tenentem et huic longitudini similem. Similiter res visa, cum fuerit ignota, conferetur magnitudo magnitudinis eius alii magnitudini rerum notarum, et adquiritur longitudo huius ymaginis per sillogismum mensure anguli quem tenet ymago in centro visus in hora reflexionis. Et locus in quo est forma rei vise comprehensus per reflexionem, forma ab eo directe veniens ad angulum circa oculum, accedet super piramidem ipsam per quam forma reflectitur ad visum, et eadem piramis occupabit totam formam que fuerit in loco ymaginis. Visus ergo, cum adquirit rem visam per reflexionem, adquirit eam in loco ymaginis, quoniam forma comprehensa in loco ymaginis per reflexionem, quare similis est forme directe comprehense, occupare ab illa piramide, et hec est causa quare comprehendatur in loco ymaginis. 
◉ Thus, if the visible object is among things familiar [to the perceiver], the visual faculty grasps its distance according to an alreadyknown distance [occupied by such a body] subtending an angle equal to this one and [lying at] a distance similar to this one. Likewise, when the visible object is unknown, its magnitude will be compared to another magnitude belonging to familiar objects, and the distance of this image is grasped according to a deduction based on the size of the angle that the image subtends at the center of sight at the time of reflection. Moreover, the place where the form of the visible object is perceived through reflection [is such that] the form reaching the eye directly from it at a [given visual] angle will arrive according to the very same cone according to which the form is reflected to the eye, and the same cone will encompass the entire form as it does at the imagelocation. Hence, when it grasps the visible object through reflection, the visual faculty grasps it where the image lies, for, because the form perceived through reflection at the imagelocation is identical to the form as perceived directly, it is contained by that [same] cone, and this is why it is perceived at the imagelocation.⁑ 
◉ Quare autem comprehendetur ymago in perpendicularem dicemus. Scimus quod punctum visui perceptibile non est intellectuale sed sensuale, et forma eius sensualis. Dico igitur in speculis planis quod ymago, cum non apparet in superficie speculi sed ultra, competentius est et rationabilius quod appareat super perpendicularem quam extra eam. Cum enim in loco perpendicularis assignata, fuerit distantia eius a puncto reflexionis speculi—que scilicet est pars linee reflexionis a loco ymaginis ad punctum reflexionis ducte—est equalis distantie puncti visi a puncto reflexionis. Quoniam superficies speculi est ortogonalis super perpendicularem, unde linea a puncto reflexionis ad perpendicularem ducta est latus duobus triangulis commune, et dividet perpendicularem per duo equalia. Quare duo latera unius trianguli erunt equalia duobus lateribus alterius, et angulus angulo, quia uterque est rectus, quare basis basi. 
◉ We will now explain why the image will be perceived on the normal. We know that a point that is perceptible to sight exists not intellectually but sensibly, and its form is sensible. I say therefore that in the case of plane mirrors, since the image does not appear on the surface of the mirror but behind it, it is more appropriate and reasonable for it to appear upon rather than outside the normal. For, since it has been defined according to a spot on the normal, the image’s distance from the point of reflection on the mirror—which is a segment of the line of reflection extended from the location of the image to the point of reflection—is equal to the distance of the point that is seen from the point of reflection. Since the surface of the mirror is orthogonal to the normal, then the line extended [on that surface] from the point of reflection to the normal forms a side common to the two triangles, and it will bisect the normal. Hence, two sides of one of the triangles will be equal to two sides of the other, and an angle [of one will be equal] to an angle [of the other], for both are right, so the base [of the one will be equal] to the base of the other.⁑ 
◉ Si ergo ymago in perpendiculari apparuerit, equaliter a puncto et a visu distabit cum corpore a quo procedit, et erit ymagini idem situs respectu puncti reflexionis qui est puncto viso respectu eiusdem, et idem est situs respectu visus, unde in hoc situ apparebit veritas et puncti visi et ymaginis. Si vero ymago fuerit extra perpendicularem, cum sit necesse eam in linea reflexionis esse, aut erit ultra perpendicularem aut citra respectu visus. Si fuerit ultra, erit quidem remotior a puncto reflexionis et a visu quam punctus visus, unde tenebit minorem angulum in oculo quam punctus. Et minorem occupat visus partem, unde, cum sit equalis, videbitur minor eo. Si autem fuerit citra perpendicularem, videbitur maior cum sit propinquior. 
◉ Thus, if the image appears on the normal, it will lie the same distance from the point [of reflection] and the eye as the object from which it originates, and the image will occupy an equivalent location with respect to the point of reflection as the point that is seen does with respect to the same point [of reflection], and the location [of both the imagepoint and objectpoint] will be equivalent with respect to the center of sight, so in this situation both the point seen and its image will appear according to reality. If, however, the image lay beyond the normal, then, since it must lie on the line of reflection, it will lie either beyond or in front of the normal with respect to the eye. If the image lies beyond, then it will lie farther from the point of reflection and from the eye than the point seen, so it will subtend a smaller angle in the eye than the point [itself]. Moreover, it occupies a smaller area on the eye, so, when it should be equal, it will appear smaller than it[s generating object]. On the other hand, if it lies in front of the normal, it will appear larger [than its generating objectpoint] because it lies nearer.⁑ 
◉ In speculo sperico exteriori videtur ymago super perpendicularem, aut enim videtur ymago centri visus, aut alterius puncti. Si ymago centri, dico quod dignior est perpendicularis ab oculo ad centrum spere ducta, ut super eam appareat ymago centri quam alia, si enim forma directe procedat secundum hanc perpendicularem usque ad centrum spere, eundem semper servabit situm respectu visus, et ita cuicumque puncto spere opponatur forma, perpendicularis ad centrum mota ydemptitatem situs tenebit respectu visus. Et idem situs erit forme in una perpendiculari qui et in alia, quoniam centrum spere eundem habet situm respectu cuiuslibet puncti spere, et omnes huiusmodi perpendiculares eiusdem sunt situs. 
◉ In the case of a convex spherical mirror, the image appears on the normal, for the image that appears is either of the center of the eye or of some other point.⁑ If the image is of the center [of the eye], then I say that it is more appropriate for the image of the center of the eye to appear upon the normal dropped from the [center of the] eye to the center of the sphere than [to appear] anywhere else, for if the form propagates directly along this normal to the center of the sphere, then it will always maintain a uniform disposition with respect to the eye, and so the form that faces any point on the sphere and moves orthogonally toward the center will maintain a uniform disposition with respect to the eye. And the disposition of the form will be the same on one normal as it is on another, for the center of the sphere has the same situation with respect to every point on the sphere, and all normals of this sort are identically situated. 
◉ Si autem extra perpendicularem ymago moveatur ad quodcumque punctum spere, mutabitur eius situs respectu visus, quoniam alium habebit situm extra perpendicularem quam in perpendiculari, et extra speculum movetur perpendicularis et non intra. Et si extra perpendicularem appareat, non servabit situm, et convenientius fuit ut servaret ymago situm quam ut mutaret, ut visus rem visam certius comprehenderet. Ob hoc ymago centri super perpendicularem apparet, et huic ymagini non possumus certum in perpendiculari assignare punctum, quoniam non invenitur dignitas in uno perpendicularis puncto plus quam in alio, ut hec ymago determinate appareat in eo. Sed scimus quod in quocumque huius perpendicularis puncto appareat semper apparet continua cum apparenti oculo, et semper in totali forma apparenti eundem tenet locum et situm. 
◉ On the other hand, if the image propagates to some point on the sphere outside the normal, its disposition with respect to the eye will change, for it will have a different situation outside the normal than it does on the normal itself, and the normal moves outside the mirror rather than inside it. But if it appears outside the normal, it will not maintain a [uniform] disposition, yet it was [agreed earlier that it is] more fitting for the image to maintain [a uniform] disposition than for it to change [its situation] if the visual faculty is to perceive the visible object with certainty. Accordingly, the image of the [eye’s] center appears on the normal, yet we cannot [on that account] determine a specific point for that image on the normal, because there is nothing to be found in any point on the normal to give it precedence over any other point in determining where the image should appear on it. But we know that at whatever point on this normal the image [of the eye’s center] appears, it invariably appears continuous with the [whole] eye appearing [in the mirror], and it always maintains [the same] location and situation with respect to the entire form [of the eye] that is seen. 
◉ Cuiuscumque puncti ymago preter centrum visus ad speculum accedat, movetur declinate, quare non durat ei similitudo situs respectu puncti visus, et perpendicularis a puncto viso ad speculum ducta cadat supra centrum spere in qua quidem perpendiculari observaret ymago similitudinem situs. Non est ergo punctus in quo comprehensa ymago servet similitudinem situs nisi in perpendiculari illa, et cum oportet ipsam comprehendi in linea reflexionis, comprehendetur in concursu linee huius cum hac perpendiculari. Iam ergo assignavimus causam huius rei, verum rerum naturalium status respicit statum suorum principiorum, et principia rerum naturalium sunt occulta. 
◉ The image of any point on the eye other than the center reaches [the mirror] obliquely, so it does not maintain a uniform disposition with respect to the objectpoint that is seen [i.e., the center of sight], and the normal dropped from the point seen to the mirror[’s surface] falls to the center of the sphere, and it is upon that normal that the image maintains a uniform situation. There is thus no point where the perceived image will maintain a uniform disposition except upon that normal, and since it must be perceived along the line of reflection, it will be perceived at the intersection of this line with that normal.⁑ We have thus explained the reason for this phenomenon, but the [essential] character of natural objects is related to the [essential] character of their [underlying] principles, and the [underlying] principles of natural objects are hidden.⁑ 
◉ Idem erit modus probationis in speculo sperico concavo; similiter in piramidali concavo vel exteriori. Et universaliter erit locus ymaginis in perpendiculari in quocumque speculo, quoniam non est locus extra perpendicularem in quo forma observet similitudinem et situs ydemptitatem. 
◉ The same way of proving this will obtain in the case of a concave spherical mirror, as well as in the case of a concave or convex conical [mirror]. In general, for any kind of mirror, imagelocation will be on the normal, for there is nowhere outside the normal where the form will maintain a uniform and equivalent disposition. 
◉ Hiis explanatis, restat demonstrative declarare locum ymaginis in qualibet speculorum specie. 
◉ With these points clarified, it remains for us to rationally define the imagelocation for every kind of mirror. 
◉ Dicimus quod quodcumque vel quodlibet punctum comprehensum a visu in speculo plano, quando egressus est a perpendiculari que a centro visus cadit in superficiem speculi plani, quod linea per quam reflectitur forma illius puncti ad visum concurret cum perpendiculari producta ab illo puncto ad superficiem speculi. Et erit punctus concursus, qui est locus ymaginis, intra speculum, et erit longitudo illius a superficie speculi equalis longitudini puncti visi a superficie speculi. Et visus non adquirit ymaginem puncti visi nisi in loco illo, et quodcumque punctum adquirit visus in hoc speculo non apparebit ex eo nisi unica ymago. 
◉ We say that, for each and every point that is perceived by sight in a plane mirror, when it lies outside the normal that falls from the center of sight to the surface of the plane mirror, the line along which the form of that point is reflected to the center of sight will intersect the normal dropped to the mirror’s surface from that point. Moreover, this intersectionpoint, which constitutes the imagelocation [of the objectpoint], lies behind the mirror[‘s surface], and this [image] will lie the same distance from the mirror’s surface as the [object]point viewed [in the mirror]. And the visual faculty perceives the image of the point viewed [in the mirror] only in that location, and only one image of any point perceived by sight in such a mirror will be seen. 
◉ Quodcumque autem punctum comprehendit visus in speculo [sperico] exteriori, quando egreditur forma perpendicularem ductam a centro visus ad centrum speculi, linea per quam reflectitur ymago ad oculum concurret cum linea producta a puncto illo ad centrum speculi, que linea est perpendicularis ducta a puncto illo ortogonalis super lineam contingentem lineam communem superficiei reflexionis et superficiei speculi. Et situs puncti concursus (qui est locus ymaginis) a superficiei speculi erit secundum situm visus a superficie speculi. Et forsitan erit punctus concursus ultra speculum, forsitan in superficie speculi, forsitan erit intra speculum. Et visus comprehendit ymagines omnes ultra speculum, licet diversa sint eorum loca, et non comprehendit locum cuiuslibet ymaginis sillogistice in superficie speculi. Et quodlibet punctum comprehensum a visu in hoc speculo non pretendit nisi unam ymaginem. 
◉ Moreover, for any point that sight perceives in a convex [spherical] mirror, when its form radiates outside the normal dropped from the center of sight to the center of the mirror, the line along which the image is reflected to the eye will intersect the line extended from that point to the center of the mirror, that line being the normal dropped from that point, and it is orthogonal to the line tangent to the common section of the plane of reflection and the surface of the mirror. Moreover, where the point of intersection (which constitutes the imagelocation) lies with respect to the mirror’s surface will depend upon where the center of sight lies with respect to the mirror’s surface. The intersectionpoint may lie beyond the mirror, on the mirror’s surface, or behind the mirror[‘s surface]. But the visual faculty perceives all [these] images behind the mirror, even though their locations may vary, and it does not perceive the location of any image on the mirror’s surface through deduction. Also, no matter what point is perceived by sight in this sort of mirror, it yields only one image.⁑ 
◉ In speculo columpnari exteriori, quodcumque punctum comprehendit visus (eadem in speculo piramidali exteriori), cum fuerit extra perpendicularem ductam a centro visus ortogonalem super superficiem contingentem superficiem speculi, linea per quam reflectitur ad visum forma concurrit cum perpendiculari ducta a puncto illo rectam super lineam contingentem lineam communem superficiei reflexionis et superficiei speculi. Et loca ymaginum horum speculorum quedam sunt ultra superficiem speculi, quedam in superficie, quedam citra. Et visus adquirit omnes ymagines horum speculorum ultra superficiem speculi, et quodcumque punctum comprehendat visus in hiis speculis non efficit nisi unam ymaginem tantum. 
◉ In the case of a convex cylindrical mirror (and the same holds for a convex conical mirror), if any given point that is perceived by sight lies outside the normal dropped from the center of sight orthogonal to the plane tangent to the mirror’s surface, the line along which the form is reflected to the center of sight intersects the normal dropped directly from that point to the line tangent to the common section of the plane of reflection and the surface of the mirror. And some of the images in these mirrors lie beyond the mirror’s surface, some on the surface itself, and some inside it.⁑ But the visual faculty grasps all of the images in these mirrors behind the mirror’s surface, and whatever point the visual faculty may perceive in these mirrors produces only one image. 
◉ In speculo sperico concavo, linee per quas reflectuntur forme punctorum visorum quedam concurrunt cum perpendicularibus ductis a punctis illis rectis super lineas contingentes lineas communes superficiei speculi et superficiei reflexionis, et quedam sunt equidistantes hiis perpendicularibus. Et que concurrunt cum perpendicularibus, locus concursus, qui est locus ymaginis, quidam ultra speculum, quidam citra speculum. Qui citra speculum fuerit quidam inter visum et speculum, quidam super centrum ipsum visus, quidam ultra centrum visus. Et adquisitio visus formarum rerum visarum quas adquirit in hiis speculis quasdam comprehendit in loco ymaginis, qui est punctus concursus—et hee sunt quas visus certe comprehendit—quasdam comprehendit extra locum concursus—et est comprehensio sine certitudine. Et res visas quas adquirit visus in hoc speculo quedam unam prefert ymaginem tantum, quedam duas, quedam tres, quedam quatuor, nec potest esse quod plures. 
◉ In the case of a concave spherical mirror, some of the lines along which the forms of visible points are reflected intersect the normals extended straight from those points to the lines tangent to the common sections of the surface of the mirror and the plane of reflection, and some are parallel to these normals. For those that intersect the normals, the point of intersection, which constitutes the imagelocation, sometimes lies behind the mirror outside the point of intersection—and this perception is unclear.⁑ Moreover, of the visible objects that sight grasps in this [sort of mirror], some yield one image only, some two, some three, and some four, but there can be no more [than four]. 
◉ In speculo piramidali concavo et columpnari concavo, linee per quas flectuntur forme ad visum quedam concurrunt in perpendicularibus ductis a punctis visis rectis super lineas contingentes lineas communes, et quedam sunt equidistantes perpendicularibus. Que concurrunt cum perpendicularibus, in quibusdam punctis concursus est ultra speculum, in quibusdam citra. Que autem citra fuerint, quedam erunt inter speculum et visum, quedam supra centrum visus, quedam ultra centrum visus. Et comprehensio rerum visarum in hoc speculo per visum, quedam sit in loco ymaginis, qui est locus concursus, quedam extra locum concursus. Et eorum que comprehenduntur aliud pretendit unam ymaginem tantum, aliud duos, aliud tres, aliud quatuor, nec aliquid est quod potest pretendere plus quam quatuor. Et nos declarabimus hec omnia demonstrative. 
◉ In a concave conical or a concave cylindrical mirror, some of the lines along which forms are reflected intersect the normals extended straight from the visible points to the lines tangent to the common sections [of the mirror’s surface and the plane of reflection], and some are parallel to the normals. In the case of those that intersect the normals, the intersection for some of the points lies behind the mirror, and the intersection for others lies in front of the mirror. Some of those that lie in front of the mirror will lie between the mirror and the eye, some at the center of sight [itself], and some beyond the center of sight. The perception by sight of visible objects in this [sort of mirror] sometimes occurs at the [proper] imagelocation, which is the point of intersection, and sometimes outside the point of intersection. And of those objects that are perceived [in this sort of mirror], one yields a single image only, another two images, another three, and another four, but there is no way that a thing can yield more than four. And we will demonstrate all these points theorematically. 
◉ [PROPOSITIO 1] Sit A [FIGURE 5.2.1, p. 563] punctus visus, B centrum visus, DGH speculum planum. Et sit G punctus reflexionis, DGH linea communis superficiei reflexionis et superficiei speculi. A puncto G ducatur EG perpendicularis super lineam communem. A puncto A ducatur perpendicularis super speculi superficiem, que sit AH, et producatur ultra speculum. Et AG sit linea per quam accedit forma ad speculum, BG per quam reflectitur ad visum. Igitur BG, EG, AG sunt in superficie reflexionis, et cum EG sit equidistans AH, et BG declinata sit super EG, concurret BG cum AH. Concurrat ergo in puncto Z. Dico quod ZH est equalis HA. 
◉ [PROPOSITION 1] Let A [figure 5.2.1, p. 220] be the objectpoint, B the center of sight, and DGH a plane mirror. Let G be the point of reflection and DGH the common section of the plane of reflection and the mirror’s surface. From point G draw EG perpendicular to that common section. From point A draw AH perpendicular to the mirror’s surface, and continue it behind the mirror. Let AG be the line along which the form [of point A] reaches the mirror and BG the line along which it is reflected to the center of sight. Accordingly, BG, EG, and AG lie in the plane of reflection, and since [normal] EG is parallel to [normal] AH, while BG is oblique to EG, BG will intersect AH. Let it then intersect at point Z. I say that ZH = HA. 
◉ Quoniam angulus BGD equalis angulo AGH, et angulus AHG equalis angulo ZHG, et latus HG commune, quare triangulus equalis triangulo; quare ZH equalis AH. 
◉ Since angle BGD = angle AGH,⁑ [since angle AGH = angle HGZ, which = vertical angle BGD], since [right] angle AHG = [right] angle ZHG, and since side HG is common, then [by Euclid, I.26] triangle [AHG] = triangle [ZHG]. Hence, ZH = AH. 
◉ Et si voluerimus per perpendicularem invenire locum reflexionis, secetur ex perpendiculari ultra speculum pars equalis parti eius usque ad speculum, et est ut sit ZH equalis AH. Et ducatur linea a centro visus ad punctum Z, que sit BGZ. Dico quod G est punctus reflexionis. 
◉ Now, if we wish to determine where the point of reflection lies according to normal [AHZ], let a segment along the normal be cut off below the mirror equal to the segment on it [from objectpoint A] to the mirror, i.e., so that ZH = AH. Then extend line BGZ from the center of sight to point Z. I say that G is the point of reflection. 
◉ Quoniam AH et HG sunt equalia HG et HZ, et angulus angulo, quare triangulus triangulo. Igitur angulus ZGH equalis angulo HGA. Sed ZGH est equalis angulo DGB. Restat ergo ut angulus BGE sit equalis angulo EGA, et ita G punctus reflexionis. Et ita propositum. 
◉ Since AH and HG = HG and HZ [leaving AH = HZ and HG common], and since [right] angle [AHG] = [right] angle [ZHG], then [by Euclid, I.4] triangle [AHG] = triangle [ZHG]. Therefore, angle ZGH = angle HGA. But [vertical angle] ZGH = [vertical] angle DGB. It follows, therefore, that angle BGE [in right angle EGD] = angle EGA [in right angle EGH], and so G is the point of reflection. And what was proposed [has] thus [been demonstrated]. 
◉ [PROPOSITIO 2] Amplius, sit A centrum visus [FIGURE 5.2.2, p. 563], et AG perpendicularis super speculum planum, et D secet hanc perpendicularem in superficie oculi. Dico quod in hac perpendiculari non est punctus qui reflectatur ab hoc speculo ad visum preter D. 
◉ [PROPOSITION 2] Furthermore, let A [figure 5.2.2, p. 220] be the center of sight, let AG be the normal to the plane mirror, and let D [be the point] on the eye’s surface [where it] intersects this normal. I say that there is no point other than D on this normal that may reflect from this mirror to the eye. 
◉ Si enim sumatur ultra visum punctus in hac perpendiculari, et sit H. Non perveniet forma eius ad speculum super perpendicularem propter solidi corporis interpositionem, et ita non reflectitur forma eius super perpendicularem. 
◉ For if [some such] point may be taken on this normal beyond the eye, let it be H. Its form will not reach the mirror along the normal because of the interference of the opaque body [of the eyeball], and so its form is not reflected along the normal. 
◉ Et si dicatur quod ab alio puncto speculi potest reflecti, sit illud B. Movebitur quidem forma eius ad punctum B per lineam HB, et reflectitur per lineam BA. Dividatur angulus HBA per equalia per lineam TB. Igitur TB erit perpendicularis super superficiem speculi. Sed TG est perpendicularis super eandem, quare ab eodem puncto est ducere duas perpendiculares ad superficiem speculi, quod est impossibile. 
◉ But if it is claimed that [its form] can be reflected from some other point on the mirror, let that [point] be B. Its form will propagate to point B along line HB, and it is reflected along line BA. Let angle HBA be bisected by line TB [so that HBT is the angle of incidence and TBA the angle of reflection]. Therefore, TB will be normal to the mirror’s surface. But TG is normal to that same [surface], so two perpendiculars will have been dropped from the same point [T] to the mirror’s surface, which is impossible. 
◉ Eadem erit probatio quod forma puncti D non potest reflecti ab alio speculi puncto quam a puncto G, quare non reflectitur nisi super perpendicularem. Punctum autem in hac perpendiculari sumptum inter G et D, si dicatur formam per reflexionem ad visum mittere improbatio, quoniam aut erit corpus solidum aut rarum. 
◉ The demonstration that the form of point D cannot be reflected from any point other than G on the mirror will be the same, so that form is reflected only along the normal. If, however, it is claimed that the form of any point selected between G and D on this normal is conveyed to the center of sight by reflection, the disproof follows from the fact that the body [of the selected point] will be either opaque or transparent. 
◉ Si solidum, procedet secundum perpendicularem forma eius ad speculum et regredietur secundum eandem usque ad ipsum, et propter soliditatem non poterit transire et ad visum pervenire. 
◉ If it is opaque, then the point’s form will proceed to the mirror along the normal and will return along the same line to the center of sight, but because of the [point’s] opacity that form cannot pass through to reach the center of sight. 
◉ Si autem punctum illud fuerit rarum, forma eius regrediens a speculo super perpendicularem miscebitur ei, et adherebit, nec reflectetur ad visum. 
◉ On the other hand, if that point is transparent, its form will mingle with it and fuse with it upon its return from the mirror along the normal, and it will not be reflected to the center of sight. 
◉ Quod autem forma cuiuscumque puncti in hac perpendiculari inter G et D sumpti non possit ab alio puncto speculi ad visum reflecti modo supradicto potest probari. Similiter, forma puncti inter A et D sumpti nec reflectitur ad visum per perpendicularem nec per aliam, quoniam puncta centrum visus et superficiem eius interposita sunt valde rara, unde nec mittitur eorum forma nec reflectitur ut sentiatur. Et quoniam quodlibet punctum preter D in superficie visus sumptum opponitur speculo non ad rectum angulum, videbitur quodlibet super perpendicularem ab eo ad speculum ductam, et eius ymago ultra speculum equidistans a superficie speculi, sicut ipsum punctum. Et quoniam D videtur continuum cum aliis superficiei visus punctis, et ymago eius continua cum aliis ymaginibus, videbitur ymago D tantum distans a superficie speculi quantum distat D ab eadem. 
◉ Moreover, by the preceding method, it can be proven that the form of any point selected on this normal between G and D cannot be reflected to the center of sight from any point on the mirror other [than G]. By the same token, the form of [any] point selected between A and D is reflected to the center of sight neither along the normal nor along any other line, for the points lying between the center of the eye and its surface are absolutely transparent, so their form is neither conveyed back nor reflected in such a way as to be sensed. And since no point other than D selected on the surface of the eye faces the mirror at a right angle, any such point will be seen upon the normal dropped from it to the mirror, and its image [will lie] the same distance behind the mirror’s surface as the point itself [lies above it]. And since D appears continuous with the other points on the surface of the eye, and since its image appears continuous with the other images [of those points], the image of D will appear [to lie] the same distance from the mirror’s surface as D lies from it.⁑ 
◉ Palam ergo quod cuiuscumque puncti in speculo visi ymago videbitur super perpendicularem, et elongatio ymaginis et visi corporis a superficie speculi eadem. 
◉ Hence, it is evident that the image of any point seen in the mirror will appear on the normal, and the distance of the image from the mirror’s surface and [the distance] of the visible body [from the mirror’s surface] is the same. 
◉ [PROPOSITIO 3] Amplius, forma puncti visi in speculo plano non reflectitur ad eundem visum nisi ab uno puncto tantum. Sit enim A [FIGURE 5.2.3, p. 563] centrum visus, B punctum visum, ZH speculum. Si ergo dicatur quod a duobus punctis speculi reflectitur forma B ad visum, sit unum punctum D, aliud E. Et ducatur linea a puncto viso ad visum, scilicet AB, que quidem linea aut erit perpendicularis supra speculum, aut non. 
◉ [PROPOSITION 3] Furthermore, the form of a point viewed in a plane mirror is reflected to the same center of sight from only one point. Let A [figure 5.2.3, p. 220] be the center of sight, B the point viewed, and ZH the mirror. If, then, it is claimed that the form of B is reflected to the center of sight from two points on the mirror, let one of them be point D, the other E. Draw line AB from the point viewed to the center of sight; this line will either be perpendicular to the mirror or not. 
◉ Si non fuerit perpendicularis, scimus quod illa linea est in superficie reflexionis ortogonali super superficiem speculi, et in una sola tali. Quoniam si in duabus, erit communis duabus superficiebus ortogonalibus, et sumpto in ea puncto, et ducta ab illo linea in alteram superficierum super lineam communem huic superficiei et superficiei speculi, erit quidem hec linea ortogonalis supra speculum. Similiter, ab eo puncto ducatur linea in alia superficie super lineam communem ei et superficiei speculi, et erit hec linea ortogonalis supra speculum, quare ab eodem puncto erit ducere duas perpendiculares. 
◉ If it is not perpendicular, we know that the line lies in the plane of reflection [which is] orthogonal to the surface of the mirror, and it will lie in only one such plane. For if [it lies] in two, it will be common to two orthogonal planes, and if a point is chosen on it and a line is extended from it in either plane to the common section of that plane and the surface of the mirror, this line will certainly be orthogonal to the mirror. Likewise, from that [same] point extend a line in the other plane to the common section of that plane and the surface of the mirror, and this line will be orthogonal to the mirror, so from the same point two perpendiculars will have been dropped [to the same plane].⁑ 
◉ Cum ergo BA sit in una sola superficie ortogonali, et tria puncta A, B, E sint in eadem superficie ortogonali, et erunt AE et EB in illa superficie ortogonali in qua est AB, similiter EB et DB. Quare EA, EB in eadem superficie cum DA, DB. Sed angulus AEH est equalis angulo DEB, et angulus HEA maior angulo ADE, quia extrinsecus, quare BED maior ADE. Sed BDZ equalis ADE, et BDZ maior BED, quare ADE maior BED, et dictum est quod minor. Restat ergo ut a solo puncto fiat reflexio. 
◉ Thus, since BA lies in only one orthogonal plane, and since the three points A, B, and E lie in the same orthogonal plane, AE and EB will also lie in the orthogonal plane that contains AB, and so will EB and DB. Therefore, EA and EB [will lie] in the same plane as DA and DB. But angle AEH = angle DEB [by supposition], and angle HEA > angle ADE, because it is an exterior [angle of triangle AED], so [angle] BED > [angle] ADE. But [angle] BDZ = [angle] ADE [by supposition], while [exterior angle] BDZ [of triangle BED] > [interior angle] BED, so [angle] ADE [which = angle BDZ, by supposition] > [angle] BED, but it was [just] claimed that it is smaller. It follows, therefore, that reflection may occur from only one point. 
◉ Si vero AB sit perpendicularis supra speculum, iam dictum est quod unicum est punctum in linea a centro visus ad speculum ortogonaliter ducta cuius forma reflectitur a speculo ad visum. Et iam probatum est quod ymago illius puncti ab uno solo reflectitur puncto, quare propositum. 
◉ On the other hand, if AB is perpendicular to the mirror, then it has already been claimed [in proposition 2] that there is only one point on the line dropped orthogonally from the center of sight to the mirror whose form is reflected from the mirror to the center of sight. And it has already been demonstrated that the image of that point reflects from only one point [on the mirror], so what was proposed [has been demonstrated]. 
◉ [PROPOSITIO 4] Amplius, inspecto aliquo puncto ab utroque visu, una tantum et eadem ymago apparet utrique, et in loco predicto. Verum planum est quod forma puncti non reflectitur ad utrumque visum ab eodem puncto speculi. Si enim linea reflexionis ad unum visum procedens angulum teneat cum perpendiculari erecta super superficiem speculi equalem angulo quem tenet linea accessus forme ad speculum cum eadem perpendiculari, non poterit in eadem superficie sumi linea alia que equalem angulum huic efficiat cum perpendiculari. Unde ab hoc puncto non reflectetur linea aliqua ad alium visum. Oportet igitur ut a diversis speculi punctis fiat reflexio. 
◉ [PROPOSITION 4] Furthermore, when any point is viewed by both eyes, one and the same image appears to both, and it does so in the aforementioned [image]location. Now, it is obvious that the form of the point does not reflect to both eyes from the same point on the mirror. For if the line of reflection proceeding to one eye were to form the same angle with the normal erected to the mirror’s surface [at the point of reflection] as the line along which the form reaches the mirror, then another line could not be chosen in the same plane to form with the normal an angle equal to this one.⁑ Hence, from this point no [form incident along the same] line will be reflected to the other eye. Reflection must therefore occur from different points on the mirror. 
◉ Sint illa puncta T, Z [FIGURE 5.2.4, p. 564]. Et sit speculum planum QE; A punctus visus; B, G duo visus; AD perpendicularis. Palam ergo quod BT et AT et ET, AD sunt in eadem superficie ortogonali super superficiem speculi. Similiter, AD, AZ, GZ sunt in eadem superficie ortogonali, et DT linea communis superficiei ADTB, et DZ communis superficiei ADZG. Si BT, GZ fuerint in eadem superficie ortogonali, erit TDZ linea una, et perpendicularis AD aut erit inter duas perpendiculares predictas ad superficiem speculi a duobus visibus, aut extra [FIGURE 5.2.4a, p. 564]. 
◉ Let those points be T and Z [figure 5.2.4, p. 221]. Let QE be the plane mirror, A the point viewed, B and G the two centers of sight, and AD the normal [dropped from objectpoint A to the mirror]. It is evident, then, that BT, AT, [segment DT of] ET, and AD lie in the same plane orthogonal to the surface of the mirror. Likewise, AD, AZ, and GZ lie in the same orthogonal plane, and DT is the common section of plane ADTB [and the mirror’s surface], while DZ is the common section of plane ADZG [and the mirror’s surface]. If BT and GZ lie in the same orthogonal plane, TDZ will form a single line, and normal AD will lie between the two aforementioned perpendiculars [dropped] to the mirror’s surface from the two centers of sight [i.e., the perpendiculars dropped from B and G to the plane of the mirror containing line QDE], or it will lie outside them [figure 5.2.4a, p. 221]. 
◉ Utrumlibet sit, linea reflexionis BT secabit ex perpendiculari AD ultra speculum partem equalem parti que est AD. Similiter, GZ secabit ex eadem perpendiculari partem ultra speculum equalem illi parti. Igitur ille due linee reflexionis secabunt perpendicularem ultra speculum in eodem puncto. Ergo ymago puncti A in eodem perpendicularis puncto percipietur ab utroque visu, quare unica tantum erit ymago et eadem, et in eodem loco que esset uno tantum visu adhibito. 
◉ Whichever the case, line BT of reflection will cut from normal AD a segment behind the mirror equal to segment AD [above it, as demonstrated in proposition 1 above]. Likewise, GZ will cut from the same normal a segment behind the mirror equal to that segment [AD above the mirror]. Those two lines of reflection will therefore intersect the normal at the same point [H] behind the mirror. Hence, the image of point A will be perceived by both eyes at the same point on the normal, so there will only be one image, and it will be the same [for both eyes], and it will lie at the same place as it would if it were viewed by only one eye. 
◉ Si vero puncta T, Z non fuerint in eadem superficie reflexionis ortogonali super speculum [FIGURE 5.2.4b, p. 564], eadem tantum erit probatio, cum utraque linea reflexionis secet ex perpendiculari partem equalem parti superiori, et erit sectio linearum reflexionis cum perpendiculari in eodem puncto, quare propositum. 
◉ If, however, points T and Z do not lie in the same plane orthogonal to the mirror [figure 5.2.4b, p. 221], precisely the same proof will apply, since each line of reflection cuts a segment on the normal [below the mirror] equal to the segment above it, and the intersection of the lines of reflection with the normal will be at the same point, so what has been proposed [is demonstrated]. 
◉ Si vero fuerit punctus A in perpendiculari ducta ab uno visu ad superficiem speculi tantum, secundum eundem visum comprehenditur ultra speculum in puncto perpendicularis tantum elongato a superficie speculi quantum distat A ab eadem, quia forma A videtur continua cum formis aliorum punctorum que quidem videntur in locis similibus. Et ab alio visu comprehenditur ymago A in eodem perpendicularis puncto, quare et sic utrique visui unica tantum apparet ymago puncti A, et in eodem eiusdem perpendicularis puncto, quod est propositum. 
◉ On the other hand, if point A lies on the normal dropped to the surface of the mirror from only one center of sight, it is perceived by that same center of sight behind the mirror at a point on the normal that lies as far from the mirror’s surface [below it] as A lies from it, so the form of A appears continuous with the forms of other points that appear in proximate locations. Also, the image of A is perceived by the other eye at the same point on the normal, and so only one image of point A is seen by both eyes, and [it appears] at the same point on the same normal, which is what was proposed.⁑ 
◉ [PROPOSITIO 5] In speculis spericis exterioribus patebit quod dicimus. Sit A [FIGURE 5.2.5, p. 565] punctus visus, B centrum visus, G punctum reflexionis. Palam quod BG, AG sunt in superficie ortogonali super superficiem contingentem speram in puncto G. Linea communis superficiei reflexionis et superficiei spere est circulus, et sit ZGQ. Linea contingens hunc circulum in puncto reflexionis sit PGE. Perpendicularis super hanc lineam sit HG. Planum quod HG perveniat ad centrum spere. Quod si non, cum linea a centro spere ducta ad punctum G sit etiam perpendicularis super lineam PGE, erit ab eodem puncto in eadem parte ducere duas lineas perpendiculares super unam lineam. 
◉ [PROPOSITION 5] What we claim will be clear in the case of spherical convex mirrors. Let A [figure 5.2.5, p. 222] be the point seen, B the center of sight, and G the point of reflection. It is clear that BG and AG lie in a plane [of reflection] orthogonal to the plane tangent to the sphere at point G. Let ZGQ be the [great] circle that forms the common section of the plane of reflection and the surface of the sphere [from which the mirror is formed]. Let PGE be the line tangent to this circle at the point of reflection. Let HG be the normal to this line [at the point of reflection]. It is evident that HG should reach the center of the circle. But if not, then, since the line extended from the center of the sphere to point G is also perpendicular to line PGE, two lines perpendicular to one [and the same] line will have been drawn from the same point on the same side [of that line].⁑ 
◉ Sit autem centrum spere N, et ducatur linea a puncto viso ad centrum spere, scilicet AN, que quidem erit perpendicularis super superficiem contingentem speram in puncto spere per quem transit. Et quoniam planum quod BG secat speram, cum sit inter HG, GP que continent rectum angulum, concurret cum linea AN. Et cum perpendicularis HG sit in superficie reflexionis, erit centrum spere in eadem, et ita AN in eadem superficie cum HG. 
◉ Now, let N be the center of the sphere, and let a line, i.e., AN, be extended from the point viewed to the center of the sphere, that line being normal to the plane tangent to the sphere at the point on the sphere through which it passes. And since it is manifest that [line of reflection] BG intersects the sphere, because it lies between HG and GP, which form a right angle, it will intersect line AN. And since normal HG lies in the plane of reflection, the center of the sphere will lie in the same plane, and so [normal] AN [will lie] in the same plane as [normal] HG. 
◉ Sit ergo concursus BG cum AN D. Planum quoniam D erit locus ymaginis, et hec quidem intelligenda sunt quando linea ducta a puncto viso ad centrum visus non fuerit perpendicularis super speculum. 
◉ Accordingly, let D be the intersection of [line of reflection] BG with [normal] AN. It is clear that D will be the imagelocation, and this analysis must be understood [to obtain] when the line extended from the point seen to the center of sight is not perpendicular to the mirror. 
◉ [PROPOSITIO 6] Amplius, linea PGE [FIGURE 5.2.6, p. 565] secat lineam AN. Sit punctus sectionis E, et dicitur punctus iste finis contingentie. Dico quod in hoc situ linea a centro spere ad locum ymaginis ducta maior est linea a loco ymaginis ducta ad locum reflexionis: id est, DN maior DG. 
◉ [PROPOSITION 6] Now, let line PGE [figure 5.2.6, p. 222] intersect line AN. Let the point of intersection be E, and this point is designated as the endpoint of tangency. I say that in this case the line extending from the center of the sphere to the imagelocation is longer than the line extending from the imagelocation to the point of reflection: that is, DN > DG. 
◉ Quoniam angulus BGH equalis angulo HGA, sed angulus BGH equalis angulo NGD. Ergo angulus HGA equalis eidem, et EG perpendicularis super HGN, quare angulus AGE equalis est angulo EGD. Igitur proportio AG ad DG sicut AE ad ED. 
◉ For angle [of reflection] BGH = angle [of incidence] HGA [by previous supposition], but angle BGH = [vertical] angle NGD. Therefore, angle [of incidence] HGA = that same [angle NGD], and EG is perpendicular to HGN, so angle AGE = angle EGD [because angle AGE = angle BGP = vertical angle EGD]. Therefore, AG:DG = AE:ED [by Euclid, VI.3]. 
◉ Protrahatur a puncto A equidistans DG, et concurrat cum linea HN in puncto H. Erit igitur angulus NGD equalis angulo GHA. Sed angulus NGD equalis est angulo AGH. Ergo angulus GHA est equalis eidem, quare duo latera AG, HA sunt equalia. Igitur proportio AH ad DG sicut AG ad idem. Sed proportio AH ad DG sicut AN ad DN, quare proportio AN ad DN sicut AG ad DG. Igitur proportio AN ad AG sicut DN ad DG. Sed AN est maior AG, quia respicit angulum maiorem recto in triangulo ANG. Igitur DN maior DG, quod est propositum. 
◉ Let a line [AH] be drawn from point A parallel to DG, and let it intersect line HN at point H. Accordingly, angle NGD = [alternate] angle GHA. But angle NGD = angle AGH [by previous conclusions]. Therefore, angle GHA = that same angle [AGH], so the two sides AG and HA [of triangle AGH] are equal. Therefore, AH:DG = AG:DG. But AH:DG = AN:DN [by Euclid, VI.4, because triangles AHN and DGN are similar], so AN:DN = AG:DG. Therefore, AN:AG = DN:DG [by Euclid, V.16]. But AN > AG, because it subtends more than a right angle in triangle ANG [by Euclid, I.19 and 32]. Accordingly, DN > DG, which is what was proposed. 
◉ [PROPOSITIO 7] Amplius, dico quod linea ducta a fine contingentie, qui est E, usque ad speram perpendiculariter, id est pars linee EN, minor est semidyametro. 
◉ [PROPOSITION 7] I say, further, that the line dropped orthogonally to the sphere from endpoint of tangency E, i.e., the segment [EF] of line EN [between E and the surface of the sphere], is shorter than the radius [of the sphere]. 
◉ Sit F [FIGURE 5.2.7, p. 566] punctus in quo AN tangit superficiem spere. Dico ergo quod EF minor est NF. 
◉ Let F [figure 5.2.7, p. 223] be the point where [normal] AN [passing through endpoint of tangency E] intersects the sphere’s surface. I say that EF < NF. 
◉ Quoniam, ut dictum est, proportio AG ad DG sicut AE ad ED, sed AN ad DN sicut AG ad GD. Igitur AN ad DN sicut AE ad ED. Igitur AN ad AE sicut DN ad DE. Sed AN maior AE, quare DN maior DE, quare DN maior EF, quare NF maior EF, quod est propositum. 
◉ For, as has [already] been claimed [in the previous proposition], AG:DG = AE:ED, but AN:DN = AG:GD [by previous conclusions]. Therefore, AN:DN = AE:ED. Therefore, AN:AE = DN:DE [by Euclid, V.16]. But AN > AE, so DN > DE, and so DN > EF, from which it follows that NF > EF, which is what was proposed.⁑ 
◉ [PROPOSITIO 8] Amplius, sit G [FIGURE 5.2.8, p. 566] centrum visus, D centrum spere, DZG perpendicularis a centro visus ad speram. Dico quod nullius puncti forma reflectitur per hanc perpendicularem nisi puncti eius quod est in superficie visus. 
◉ [PROPOSITION 8] Now, let G [figure 5.2.8, p. 223] be the center of sight, D the center of the sphere, and DZG the normal [dropped] from the center of sight to the sphere. I say that the form of no point other than the point on the surface of the eye [through which normal GD passes] is reflected along this normal. 
◉ Punctorum enim forme post centrum visus sumptorum non reflectuntur per eam propter causam supradictam. Similiter nec puncta inter superficiem visus et speculum sumpta. Dico etiam quod nullum punctum huius perpendicularis reflectitur ab alio puncto speculi. 
◉ For the forms of the points selected beyond the center of the eye are not reflected for the reason given above [in proposition 2]. And the same holds for the points lying between the surface of the eye and the mirror. I say, as well, that no point on this normal is reflected from any other point on the mirror. 
◉ Si enim dicatur quod ab alio puncto, sit illud punctum A. Erit linea GA linea reflexionis, et a puncto illo intelligemus lineam ad A, que est linea per quam movetur forma. Et includunt hee due linee angulum super A, quem quidem angulum necessario dividet dyameter DA, cum sit perpendicularis super punctum A, quia perpendicularis dividit angulum ex linea motus forme et linea reflexionis per equa. Et ita dyameter DA concurret cum perpendiculari GD inter punctum sumptum et G. Et ita due linee recte in duobus punctis concurrent et superficiem includent. 
◉ For if it is claimed that [reflection does occur] from another point, let that point be A. Line GA will be the line of reflection, and from that point [on the normal, which is assumed to reflect its form to G, i.e., X] we will imagine a line to A, that line [AX] being the one along which the form proceeds [to the mirror]. These two lines form an angle at A, and diameter DA will necessarily divide that angle, since it is normal to point A, and the normal bisects the angle produced by the form’s line of incidence and its line of reflection. And so diameter DA will intersect normal GD between the selected point and G [i.e., at point E]. And so the two straight lines [GD and DE, which consists of DA and its rectilinear continuation AE] will intersect at two points [D and E] and will form a plane. 
◉ Restat ergo ut solius puncti qui est in superficie visus forma reflectatur a speculo ad perpendicularem, et videatur in primo ymaginis loco propter eius cum aliis continuitatem. 
◉ It therefore follows [from the impossibility of such a double intersection] that the form of only the one point that lies on the surface of the eye may be reflected from the mirror along the normal, and it must appear at its original imagelocation according to its continuity with other [surrounding] points.⁑ 
◉ [PROPOSITIO 9] Amplius, GA, GB [FIGURE 5.2.9, p. 567] sint linee a centro visus ducte contingentes speram, et signetur circulus super quem superficies hiis lineis inclusa secat speram. Erit AB portio apparens ex hoc circulo. Dico ergo quod loca ymaginum que per reflexiones ab hac portione factas comprehenduntur quedam sunt intra speculum, quedam in superficie speculi, quedam extra speculum. Et singulum horum est determinandum. 
◉ [PROPOSITION 9] Furthermore, let GA and GB [figure 5.2.9, p. 224] be lines drawn tangent to the sphere from the center of sight, and mark off the [great] circle upon which the plane formed by these lines cuts the sphere. AB will be the visible portion of this circle. I say, therefore, that some of the imagelocations perceived according to reflection from this portion lie inside the mirror, some on the mirror’s surface, and some outside the mirror. Each one of these cases must be accounted for. 
◉ Ducatur a puncto G linea secans circulum, et pars eius que est corda arcus circuli sit equalis semidyametro circuli. Sit linea illa GHK, et corda equalis semidyametro sit HK. Et producatur a puncto H perpendicularis, que sit DHM. Dico quod forma reflexa a puncto H locus eius erit intra speram. 
◉ From point G [figure 5.2.9] draw a line that cuts the circle, and let the segment of it that forms the chord on the [intersected] arc of the circle be equal to the circle’s radius. Let that [cutting] line be GHK, and let HK be the [segment on it forming the] chord that is equal to the radius. Then, from point H draw normal DHM. I say that the [image]location of a form reflected from point H will lie inside the sphere. 
◉ Ducatur a puncto H linea equalem tenens angulum cum MH angulo MHG, et sit OH. Reflectentur quidem puncta huius linee a puncto H ad visum, et non alterius. Sumatur ergo aliquod eius punctum, et sit O, et ducatur ab eo linea ad centrum spere, que sit OD. Erit quidem OD perpendicularis super superficiem contingentem speram super punctum eius per quod transit OD. Verum angulus OHM equalis est angulo, ex ypothesi, MHG, quare similiter equalis est angulo contraposito, scilicet KHD. Sed KHD est equalis KDH, quoniam respiciunt equalia latera. 
◉ From point H draw a line OH that forms with MH an angle [OHM] equal to angle MHG. Points [incident to the mirror] on this line, and on no other, will be reflected from point H to the center of sight [G]. Accordingly, choose some point O on it, and draw line OD [i.e., the normal dropped from the objectpoint] from that point to the center of the sphere. OD will be perpendicular to the plane tangent to the sphere at the point where OD passes through it. But, by construction, angle OHM = angle MHG, so, by the same token, it is equal to [angle MHG’s] alternate angle KHD. But [angle] KHD = [angle] KDH, because they are subtended by equal sides [of equilateral triangle KDH, i.e., radii DH and DK]. 
◉ Igitur angulus OHM equalis est angulo KDM, quare linee KD, OH sunt equidistantes. Ergo in infinitum producte numquam concurrent. Et linea OD secabit lineam interiacentem KD, OH, et ita quodcumque punctum sumatur in linea OH, linea ducta ab illo puncto ad punctum D secabit lineam reflexionis intra speram, que quidem linea erit perpendicularis super speram, sicut est OD. Quare ymago cuiuscumque puncti linee OH apparebit intra speram. 
◉ Accordingly, angle OHM = angle KDM, so lines KD and OH are parallel. Therefore, if they are extended indefinitely, they will never intersect.⁑ But line OD will intersect the line [HK] joining KD and OH, and so, no matter what point is chosen on line OH, the line extended from that point to point D [i.e., the normal] will intersect the line of reflection inside the sphere, and that line will be perpendicular to the sphere, as represented by OD [so it will be the normal on which the image must lie]. Hence, the image of any point on line OH will appear inside the sphere [where the line of reflection GH intersects normal OD]. 
◉ Amplius, arcus circuli interiacens punctum H et punctum per quem transit perpendicularis a centro visus ducta est HZ. Dico quod, a quocumque puncto huius arcus fiat reflexio, locus ymaginis erit intra speram. 
◉ Now, the arc on the circle lying between point H and the point where the normal dropped from the center of sight passes through [the circle] is HZ. I say that, from whatever point on this arc reflection occurs, the imagelocation will lie inside the sphere. 
◉ Probatio: sit I [FIGURE 5.2.9a, p. 567] punctus sumptus, et ducatur linea a centro visus secans circulum super punctum illum, que sit GIS, et ducatur perpendicularis a puncto hoc que sit DIT. Et fiat linea PI equalem tenens angulum cum IT angulo TIG. Palam quod sola puncta linee PI reflectuntur a puncto I ad visum. Palam etiam quod linea IS maior est linea KH, quare maior SD. Igitur angulus SDI maior est angulo SID, quare est maior angulo GIT, quare maior angulo TIP. 
◉ [Here is] the proof. Let I [figure 5.2.9a, p. 224] be the point selected, draw line GIS from the center of sight to intersect the circle at that point, and draw normal DIT from that point. Make line PI form an angle [PIT] with [normal] IT that is equal to angle TIG. It is clear that only points on line PI are reflected from point I to the center of sight. It is also clear that line IS > line KH [by Euclid, III.8], so it is longer than [line] SD [which = line KH, by construction]. Therefore, angle SDI > angle SID [by Euclid, I.19] , so it is greater than angle GIT [which is vertical to angle SID], and therefore greater than angle TIP. 
◉ Igitur linea PI et SD numquam concurrent, et linea ducta a puncto quocumque PI linee ad punctum D secabit lineam SI intra speram, que SI est linea reflexionis. Et omnis linea ducta a quocumque puncto PI linee erit perpendicularis super speram, sicut est PD. Et cum locus ymaginis sit in concursu perpendicularis a puncto viso et linee reflexionis, erit ymago cuiuslibet puncti linee PI intra speram. Palam ergo quod omnium ymaginum arcus HZ locus proprius erit intra speculum, quod est propositum. 
◉ Hence, line PI and [line] SD will never intersect [in the direction of P and S],⁑ and the line extended to point D from any point on line PI will intersect line SI inside the sphere, SI being the line of reflection [continued inside the circle]. And any line drawn [to D] from any point on line PI will be perpendicular to the sphere, just as PD is. Moreover, since the imagelocation lies at the intersection of the normal [dropped] from the point viewed and the line of reflection, the image of any point on line PI will lie inside the sphere. It is therefore evident that the appropriate location of all images [produced by reflection from points] within arc HZ will be inside the sphere, which is what was proposed. 
◉ Amplius, sumpto quocumque puncto arcus HB, dico quod quedam eius ymago erit intra speculum, quedam in superficie speculi, quedam extra speculum. 
◉ Moroever, if some point [of reflection] is chosen on arc HB, I say that the image will sometimes lie inside the sphere, sometimes on the surface of the sphere, and sometimes outside the sphere. 
◉ Sumatur aliquod eius punctum, et sit N [FIGURE 5.2.9b, p. 567], et ducatur linea a puncto G secans circulum, que sit GNQ. Et ducatur perpendicularis DNF, et protrahatur linea equalem angulum tenens cum perpendiculari angulo FNG, et sit EN. Quoniam linea NQ minor est linea KH, est etiam minor linea QD, et ita angulus QDN minor est angulo DNQ, quare minor angulo GNF, quare etiam minor angulo ENF. Igitur linea EN et DQ concurrent. Sit ergo concursus in puncto E. Palam quod linea EQD est perpendicularis super speram, et secat lineam GNQ, que est linea reflexionis, in puncto Q, qui est punctus spere. Quare ymago puncti E, cum fuerit reflexio super punctum N, apparebit in puncto Q, et est in superficie spere. 
◉ Choose some point on this arc, let it be N [figure 5.2.9b, p. 224], and from point G draw line GNQ to intersect the circle. Then draw normal DNF, and draw line EN to form with normal [DNF] an angle [ENF] equal to angle FNG. Since line NQ < line KH [by Euclid, III.8], it is also shorter than line QD [since QD = KH, by construction], and so angle QDN < angle DNQ [by Euclid, I.19], and therefore smaller than angle GNF [which is vertical to angle DNQ], and therefore also smaller than angle ENF [which = angle GNF, by construction]. Hence, line EN and [line] DQ will intersect. So let the intersection be at point E. It is obvious that line EQD is normal to the sphere, and it intersects line of reflection GNQ at point Q, which is a point on the sphere. Therefore, if reflection occurs at point N, the image of point E will appear at point Q, and this point lies on the surface of the sphere. 
◉ Si vero in linea EN sumatur punctum ultra E, utpote R, perpendicularis ducta ab eo ad centrum spere, sicut RD, secabit lineam reflexionis, que est GNQ, ultra punctum Q. Et est extra speram, quare ymago cuiuslibet puncti linee EN ultra E sumpti erit extra superficiem speculi. 
◉ If, however, a point, such as R, is chosen on line EN beyond E, the normal dropped from that point to the center of the sphere, e.g., RD, will intersect line of reflection GNQ beyond point Q [e.g., at point L]. This lies outside the sphere, so the image of any point chosen on line EN beyond E will lie outside the surface of the mirror. 
◉ Si vero in linea EN citra punctum E sumatur aliquod punctum, perpendicularis ab eo ducta ad speculum secabit lineam GNQ intra speram, quoniam in puncto quod sit inter N et Q. Quare ymago cuiuslibet puncti linee EN inter E et N sumpti apparebit intra speram. 
◉ But if some point [such as C] is chosen on line EN in front of point E, the normal dropped from it to the mirror [along CD] will intersect line GNQ inside the sphere, because [it intersects it] at a point that lies between N and Q. Accordingly, the image of any point chosen on line EN between E and N will appear inside the sphere. 
◉ Eadem penitus erit probatio, sumpto quocumque alio arcus BH puncto. Et ita ymago cuiuslibet puncti arcus BH una sola est in superficie speculi; aliarum quedam in speculo, quedam extra. Et quod demonstratum est in arcu ZB eodem modo potest patere in arcu ZA, et eadem penitus erit demonstratio, cuiuscumque circuli spere sumatur portio visui opposita et perpendiculari GD equaliter divisa. 
◉ The very same proof will hold for any other point chosen on arc BH. And so, for any point on arc BH, one image alone lies on the surface of the mirror, others lie inside the mirror, and others lie outside it. Moreover, what has been demonstrated for arc ZB can be shown in exactly the same way for arc ZA, and the very same demonstration will apply for any [great] circle on the sphere and the portion [of it] that is selected to face the center of sight and be bisected by normal GD. 
◉ Unde visu immoto et perpendiculari GD manente, si moveatur equidistans perpendiculari linea GHK, secabit ex spera motu suo portionem circularem, et cuiuslibet puncti huius portionis ymago apparebit intra speram. 
◉ Accordingly, if the center of sight remains immobile with normal GD fixed in place, and if line GHK is rotated uniformly about normal [GD as axis], it will cut off a circular portion of the sphere with its rotation, and the image of any point [reflected from any point] within this portion will appear inside the sphere.⁑ 
◉ Si vero linea contingens GB moveatur equidistanter perpendiculari visus, secabit ex spera portionem predictam maiorem, et a quolibet puncto excrementi unius super aliam refertur ymago cuius locus erit in superficie spere; et aliarum quedam intra speram, quedam extra. 
◉ On the other hand, if line GB tangent [to the sphere] is rotated uniformly about normal [GD as axis], it will cut off a portion of the sphere larger than the previous one, and from any point on the portion that constitutes the excess of the latter over the former, [one] image reflected [to the eye] will find its location on the surface of the sphere, and of the remaining images, some [will lie] inside the sphere, and some outside it.⁑ 
◉ Scimus ex hiis quod in hoc speculo quelibet ymago apparet in dyametro spere, aut intra, aut extra, aut in superficie. Et omnis dyameter in quo appareat ymago aliqua in superficie spere aut extra dimissior est puncto spere quem tangit linea contingens a centro visus ducta, id est ultimo puncto portionis apparentis. Scimus quod quelibet linea reflexionis secat speram in duobus punctis: in puncto reflexionis et in alio. 
◉ From these facts, we know that in this [sort of] mirror any image appears on a diameter of the sphere [that forms the normal, and that image may appear] inside the sphere, outside it, or on its surface. Moreover, any diameter upon which an image may appear, whether on the surface of the sphere or outside that surface, lies below the point on the sphere where the tangent extended from the center of sight to the sphere touches the sphere, that point being at the very edge of the visible portion [of the mirror]. We [also] know [from this analysis] that every line of reflection cuts the sphere in two points, the point of reflection and some other one.⁑ 
◉ [PROPOSITIO 10] Restat ut loca ymaginum certius determinemus. Dico quod, sumpto dyametro, si ad ipsum ducatur linea secans speram a centro visus cuius pars interiacens punctum sectionis spere et punctum dyametri quem attingit est equalis parti dyametri interiacenti punctum illud et centrum, punctus ille non est locus alicuius ymaginis. 
◉ [PROPOSITION 10] It remains for us to determine imagelocations more specifically. I say that, if a diameter is chosen [on a convex spherical mirror], and if a line is extended to it from the center of sight to cut the sphere so that the segment lying between the point of intersection on the sphere and the point on the diameter to which it reaches is equal to the segment of the diameter lying between that latter point and the center [of the sphere], that point does not constitute the location of any image. 
◉ Verbi gratia, sit AG [FIGURE 5.2.10, p. 568] circulus spere, H visus, ED dyameter spere, sive perpendicularis. Et HZ sit linea secans speram super punctum F et concurrens cum ED in puncto Z, et sit ZF equalis ZD. Dico quod Z non est locus alicuius ymaginis. 
◉ For instance, let AG [figure 5.2.10, p. 225] be a [great] circle on the sphere, H the center of sight, and ED a diameter, or normal, in the sphere. Let HZ be a line intersecting the sphere at point F and meeting ED at point Z, and let ZF = ZD. I say that Z does not constitute the location of any image [on normal DE]. 
◉ Palam enim quod non est locus ymaginis alterius quam alicuius puncti ED, quoniam ymago cuiuslibet puncti est super dyametrum ab eo ad centrum spere ductum. Et quod locus ymaginis alicuius puncti ED non sit in Z sic constabit. 
◉ For it is evident that there is no imagelocation at any point other than on ED, because the image of any point lies on the diameter extended from it to the center of the sphere. That the imagelocation of any point on ED does not lie at Z will be demonstrated as follows. 
◉ Ducatur perpendicularis a puncto D super punctum F, que sit DFN, et super punctum F fiat angulus equalis angulo NFH, et sit QFN. Palam igitur quod angulus QFN est equalis angulo ZFD. Sed ZFD est equalis angulo ZDF, quia respiciunt equalia latera. Igitur QFN est equalis angulo ZDN, quare linea FQ est equidistans linee ED. 
◉ From point D draw normal DFN to point [of reflection] F, and at point F form angle QFN equal to angle NFH. It is therefore evident that angle QFN = angle ZFD [which = vertical angle NFH]. But [angle] ZFD = angle ZDF, because they are subtended by equal sides [of isosceles triangle ZFD, sides ZF and ZD being equal, by construction]. Therefore, [angle] QFN = angle ZDN, so [by Euclid, I.27] line [of incidence] FQ is parallel to line ED [which forms the normal]. 
◉ Igitur in infinitum producte numquam concurrent. Igitur nullius puncti ED forma movebitur ad punctum F per QF, et non potest esse locus ymaginis alicuius puncti in puncto Z nisi forma eius moveatur ad F per lineam QF. Eadem erit improbatio, sumpto quocumque dyametro, quare propositum. 
◉ Hence, if they are extended indefinitely, they will never meet. Accordingly, the form of no point on [normal] ED will pass to point F along [line of incidence] QF, and there can be no imagelocation for any point at point Z unless its form passes to F along line QF.⁑ The same disproof will hold for any diameter that is chosen, so what was proposed [is demonstrated]. 
◉ Amplius, dico quod nullus punctus linee ZD potest esse locus alicuius ymaginis. [2.102] Sumatur enim punctus P, et ducatur linea HP secans speram in puncto B. Et ducatur perpendicularis DBM, et angulo MBH fiat angulus equalis, qui sit TBM. Palam quod TBM est equalis PBD, et palam quod angulus DPH est maior angulo PZF, quia extrinsecus. Igitur duo alii anguli trianguli DPB sunt minores duobus aliis angulis trianguli DZF. Sed angulus PDB maior angulo ZDF. Restat ergo ut angulus DBP sit minor angulo DFZ. Sed angulus DFZ est equalis angulo ZDF, quare angulus DPB minor est angulo ZDF. Igitur multo minor angulo PDB, quare angulus TBM minor est angulo PDB. Igitur linee TB, ED numquam concurrent, et ita nulla ymago puncti B refertur ad punctum P. Similiter, nec ymago alterius puncti, et similiter de quolibet puncto linee ZD. Restat ergo ut tota ZD sit vacua a locis ymaginum. 
◉ I say, as well, that no point on line ZD can be the location for any image. 
◉ [PROPOSITIO 11] Amplius, sumpto quocumque dyametro inter lineas contingentie a visu ad speram ductas, preter dyametrum a centro visus ad centrum spere intellectum, et determinato in eo puncto quem diximus, qui est meta locorum ymaginum, dico quod in punctis tantum illius dyametri qui sunt inter superficiem spere et metam predictam sunt loca ymaginum punctorum illius dyametri. 
◉ [PROPOSITION 11] Furthermore, if a diameter, other than the one imagined [to extend] from the center of sight to the center of the sphere, is taken between the lines drawn tangent to the sphere from the center of sight, and if the point on it we [just] described as the limit of imagelocations is determined, I say that the imagelocations of points on that diameter lie only at points on it between the surface of the sphere and the aforementioned limit. 
◉ Verbi gratia, sint BZ, BE [FIGURE 5.2.11, p. 568] linee contingentie, B centrum visus, A centrum spere, BHA dyameter visualis, DA dyameter sumptus cum meta G punctus spere. Dico quod in sola puncta G et T interiacentia cadunt ymagines punctorum DA. 
◉ For example, let BZ and BE [figure 5.2.11, p. 225] be the tangents, B the center of sight, A the center of the sphere, BHA the visual axis [which is normal to the mirror from the center of sight], and DA the selected diameter with G [the point at which DA intersects the sphere and T] the limit [of imagelocations].⁑ I say that images of points [whose forms are reflected to B] occur only at points between G and T on DA. 
◉ Quod enim non cadent in puncto G vel extra superficiem spere palam per hoc quod supradictum est dyametrum in quo locus ymaginis erit in superficie spere aut extra demissiorem esse puncto contingentie; et cum dyameter DA sit inter lineas contingentie, non erit in eo locus ymaginis, aut in superficie spere aut extra. Quod autem in quodlibet punctum inter G et T sumptum cadat ymago constabit. 
◉ That they will not occur at point G or [any point] outside the sphere’s surface is clear from what was said above [in proposition 9], i.e., that the diameter on which the imagelocation will occur at the surface of the mirror or outside it lies below the point of tangency [E]; and since diameter DA lies between the lines of tangency, there will be no imagelocation on it either at or outside the [mirror’s] surface. On the other hand, it will be demonstrated that the image may fall at any point taken between G and T. 
◉ Sumatur punctum, et sit Q, et ducatur linea BQ secans speram in puncto C. Et ducatur perpendicularis ACL, et angulo LCB fiat angulus equalis DCL, et educatur linea BT secans speram in puncto F, et ducatur perpendicularis AF. Igitur triangulus ACB continet triangulum AFB, quare angulus AFB maior angulo ACB. Restat ergo ut angulus AFT sit minor ACQ. Sed angulus AFT est equalis angulo FAT, quia equalia latera respiciunt. Igitur ACQ maior erit angulo CAQ, quare LCB maior CAQ, unde DCL maior CAQ. Igitur CD, AQ concurrent. Sit D concursus. Forma igitur puncti D reflectatur in puncto C per lineam CB, et locus ymaginis eius est Q. Et eadem est probatio, sumpto quocumque puncto inter G et T. 
◉ Choose [such] a point, let it be Q, and draw line [of reflection] BQ to cut the sphere at point C. Now, draw normal ACL, make angle [of incidence] DCL = angle [of reflection] LCB, draw line BT to cut the sphere at point F, and draw normal AF. Accordingly, triangle ACB contains triangle AFB, so angle AFB > angle ACB. It follows, then, that angle AFT [adjacent to larger angle AFB] < [angle] ACQ [adjacent to smaller angle ACB]. But angle AFT = angle FAT, because they are subtended by equal sides [of isosceles triangle FAT]. Therefore [angle] ACQ > angle CAQ [because angle CAQ < angle FAT < angle ACQ], so [angle] LCB [which = vertical angle ACQ] > [angle] CAQ, from which it follows that [angle] DCL > [angle] CAQ [since angle DCL = angle LCB, by construction]. CD and AQ will therefore intersect. Let D be the [point of] intersection. Thus, the form of point D should reflect at point C along line CB, and its imagelocation is Q. And the same proof holds for any [other] point selected between G and T. 
◉ [PROPOSITIO 12] Restat ut assignemus loca ymaginum in sectione spere occulta visui. 
◉ [PROPOSITION 12] It remains for us to determine imagelocations in the invisible section of the sphere. 
◉ Sint ergo AC, AG [FIGURE 5.2.12, p. 568] linee contingentes portionem apparentem, A centrum visus, B centrum spere, ADBZ dyameter visualis, et ZCG circulus spere in superficie linearum contingentie. Et protrahatur a centro ad punctum contingentie dyameter BG. Palam quod angulus ZBG est maior recto, cum enim in triangulo BAG angulus BGA sit rectus, erit angulus GBA minor recto, quare ZBG maior. Sit ergo HBG rectus. Erit ergo HB equidistans linee contingentie AG. Igitur producte numquam concurrent, et quilibet dyameter inter H et G concurret cum linea AG. 
◉ Accordingly, let AC and AG [figure 5.2.12, p. 225] be the tangents [defining] the visible portion [of the sphere], A the center of sight, B the center of the sphere, ADBZ the visual axis [which is normal to the mirror from the center of sight], and ZCG the [great] circle on the sphere lying in the plane of the tangent lines. From the center [of the sphere] draw diameter BG to point [G] of tangency. It is evident that angle ZBG > a right angle, for, since angle BGA in triangle BAG is right, angle GBA < a right angle, so [adjacent angle] ZBG will be greater [than a right angle]. Accordingly, let HBG be a right angle. HB will thus be parallel to line of tangency AG. Hence, when extended [indefinitely] they will never intersect, whereas any diameter between H and G will intersect line AG. 
◉ Ducatur a puncto A linea secans speram, que sit AMO, ita quod corda, que est MO, sit equalis semidyametro OB, et concurrat dyametro BO cum linea AG in puncto T. Dico quod in quolibet puncto TO est locus ymaginis, et in nullo alio puncto dyametri TB est locus ymaginis, et sunt O, T termini locorum ymaginis. 
◉ Draw line AMO from point A to intersect the sphere so that chord MO = radius OB, and let diameter BO intersect line AG at point T. I say that there is an imagelocation at any point on TO, that there is an imagelocation at no other point on diameter TB, and that O and T are the limitpoints for imagelocations [on that diameter].⁑ 
◉ Sumatur enim punctum, et sit K, et ducatur ANK secans speram in puncto N. Et ducatur perpendicularis BNX, et angulo XNA fiat angulus equalis per lineam FN. Palam quod FN non cadet inter B et T, quoniam aut secans speram aut secat contingentem AP in duobus punctis. Igitur forma puncti F movebitur per FN ad punctum N, et reflectetur ad A per lineam AN, et apparebit ymago eius in puncto K. Et eadem probatio, sumpto quocumque alio puncto. 
◉ Take [such a] point, let it be K, and draw [line of reflection] ANK to cut the sphere at point N. Then draw normal BNY, and form angle [of incidence FNY] with FN that is equal to angle [of reflection] YNA. It is clear that [line of incidence] FN will not lie between B and T, because [in that case] it must cut either the sphere or tangent AP in two points.⁑ Therefore, the form of point F will propagate along FN to point N and will be reflected to A along line AN, and its image will appear at point K. And the same proof holds for any other point chosen [as an imagelocation between T and O]. 
◉ [PROPOSITIO 13] Amplius, dico quod in arcu OG, quicumque sumatur dyameter, continebit loca ymaginum et intra speculum, et unam in superficie speculi, et alias extra speculum. 
◉ [PROPOSITION 13] I say, further, that no matter what diameter may be chosen within arc OG, it will contain imagelocations inside the mirror, as well as one [image] on the surface of the mirror, and other [images] outside the mirror. 
◉ Sumatur ergo punctus L [FIGURE 5.2.13, p. 569], et protrahatur dyameter BL usquoque secet AP in puncto E. Et producatur linea AL secans speram in puncto R. Palam quod RL minor est LB, quia est minor MO, que est equalis semidyametro. Si ergo ab A ducatur linea ad dyametrum LB cuius pars interiacens circulum et dyameter sit equalis parti dyametri a puncto in quod cadit ad centrum, cadet quidem inter L et B. Si enim inter L et E ceciderit, erit RL maior LB, et omnis linea interiacens centrum et illam equalem erit maior parte dyametri que terminatur, secundum probationem assignatam in explanatione mete ymaginum. 
◉ Accordingly, take point L [in figure 5.2.13, p. 226], and draw diameter BL until it intersects [line of tangency] AP at point E. Then draw line AL to intersect the sphere at point R. It is obvious [by Euclid, III.8] that RL < LB, since it is shorter than MO, which is equal to the [sphere’s] radius [by construction from the previous proposition]. Accordingly, if from A a line is drawn to diameter LB such that the segment [of it] lying between the circle and the diameter is equal to the segment of the diameter [extending] from the point where that line meets the diameter to the center [of the circle], it will fall between L and B. For if it were to fall between L and E, then RL > LB, and every line falling between the center [of the sphere] and that equal segment [on LB] would be longer than the segment on the diameter where it terminates, according to the proof [in proposition 10] devoted to explaining the limit of image[locations].⁑ 
◉ Sit ergo punctus in quem linea equalis cadet I. Dico quod in quolibet puncto in EI sumpto est locus ymaginis. Et erit eadem demonstratio que fuit in TO. 
◉ So let I be the point where the line[segment between arc DG and normal BL] equal [to the segment on BL] will fall. I say that there is an imagelocation at any point on EI [other than E or I]. And the same demonstration will apply [in this case] as applied for TO [in the previous theorem]. 
◉ Igitur quedam ymagines in dyametro EB sortiuntur loca intra speculum, quedam extra speculum, una sola in superficie, scilicet puncto L. Et ita poteris demonstrare in quolibet dyametro partium OG transeunte. 
◉ Accordingly, the imagelocations on diameter EB are divided among those that lie inside the mirror [i.e., between L and I], those that lie outside the mirror [i.e., between E and L], and the single one, i.e., at point L, that lies on the mirror’s surface. And in this way you can demonstrate [the same thing] for any diameter that passes through arc OG. 
◉ [PROPOSITIO 14] Amplius, sumpto quocumque dyametro in arcu OH, locus ymaginis in eo erit extra speculum. 
◉ [PROPOSITION 14] Moreover, if some diameter is taken within arc OH, [any] imagelocation on that diameter will lie outside the mirror. 
◉ Sumatur dyameter BQ [FIGURE 5.2.14, p. 570], et concurrat cum contingente in puncto P. Et ducatur linea ANQ secans speram in puncto N. Iam dictum est quod MO est equalis OB, sed NQ est maior MO, quare NQ est maior QB. Et linea ducta ad circumferentiam ad dyametrum PB equalis parti BP ipsam et centrum interiacenti non cadet inter Q et B. Si enim ceciderit, secundum supradictam probationem erit NQ minor QB. 
◉ Take diameter BQ [figure 5.2.14, p. 227], and let it intersect the tangent [AG] at point P. Then draw line ANQ to cut the sphere at point N. It has already been established [by construction in propositions 12 and 13] that MO = [radius] OB, but NQ > MO, so NQ > [radius] QB. Moreover, the line drawn at the circumference [of the sphere] to diameter PB and equal to the segment of BP lying between it and the center [of the sphere] will not fall between Q and B. For if it were to fall [there], then, according to the previous demonstration, NQ < QB [which contradicts the conclusion drawn above]. 
◉ Restat ergo ut linea equalis cadat inter P et Q. Et quod non cadat in punctum P palam per hoc quod angulus PGB rectus. Igitur PB maius PG. Cadet ergo citra P. 
◉ It follows, therefore, that the line equal [to the previously defined line] should fall between P and Q. That it may not fall at point P is clear from the fact that angle PGB is a right angle [and so tangent AGP cannot be a line of reflection]. Therefore, PB > PG [because it subtends a larger angle in triangle BGP]. So it will fall in front of P [i.e., between P and the mirror]. 
◉ Sit punctus in quem cadit S. Erit igitur S meta locorum ymaginum, et quilibet punctus inter P et S erit locus ymaginum, et est eadem probatio quam supra. 
◉ Let the point at which it falls be S [so that the linesegment on AS between S and the point at which AS cuts the circle is equal to SB]. S will therefore constitute the limit for imagelocations, and any point between P and S will constitute an imagelocation, and the proof for this is the same as above. 
◉ Palam ex hiis quod ymagines dyametrorum arcus HO omnes sunt extra; ymagines dyametri FB una in superficie, que est in O, alie omnes extra, scilicet in TO; omnes autem ymagines dyametri arcus OG, quedam intra, quedam extra, quedam in superficie. 
◉ From these theorems it is evident that all the images [produced] on diameters within arc HO lie outside [the mirror]; on diameter FB, one lies on the surface, at O; and all the rest lie beyond, i.e., in TO; however, of all the images on [any] diameter within arc OG, some lie inside [the mirror], some outside it, and one on its surface. 
◉ [PROPOSITIO 15] Amplius, in arcu HZ [FIGURE 5.2.15, p. 570] non potest sumi dyameter in quo est locus ymaginis, quoniam nullus dyameter ibi sumptus concurrit cum contingente AP. 
◉ [PROPOSITION 15] Furthermore, no diameter can be selected in arc HZ [figure 5.2.15, p. 227] that contains an imagelocation, for no diameter selected there intersects tangent AP [since HB is parallel to AP]. 
◉ Et a quocumque puncto illius talis dyametri ducatur linea ad speram. Cadet quidem in portione GZC et nulla in portione GDC, nisi secando speram. Ergo nulla forma puncti alicuius talis dyametri veniet ad portionem visui apparentem. 
◉ Now, from some point on such a diameter [outside the sphere], draw a line [tangent] to the sphere. It will of course touch [the sphere] within segment GZC, not within segment GDC, unless it cuts the sphere. Accordingly, no form of any point on such a diameter will reach the portion [of the mirror] visible to the eye. 
◉ Quod autem dictum in arcu GH potest eodem modo demonstrari in parte arcus CZ eam respiciente. Et sumpto arcu citra Z equali HZ, in nullo dyametro illius arcus erit ymaginis locus. 
◉ Moreover, what has been claimed for arc GH can be demonstrated in the same way for the portion of arc CZ that corresponds to it. So if an arc equal to arc HZ is taken on the other side of Z, there will be no imagelocation on any diameter within that arc. 
◉ Idem est demonstrandi modus in quocumque circulo, quare, si linea HB moveatur, eodem manente angulo HBZ, signabit motu suo portionem spere in dyametris cuius nullus sit ymaginis locus. Si vero HB immota, moveatur OB, describetur portio cuius omnes ymagines extra; verum dyametrum TB una in superficie, alie extra. Moto autem arcu OG, fiet portio cuius quedam ymagines in superficie, quedam extra speculum, quedam intra. 
◉ The same method is used in proving this for any circle [on the sphere], so if line HB is rotated [about axis BZ] while angle HBZ is kept constant [throughout its rotation], it will describe with its motion a portion of the sphere within which there is no diameter with an imagelocation. If, however, HB is held in place [so that angle HBZ remains constant], and if OB is rotated [about axis BZ], it will describe a portion [of the sphere] within which all of the images [on any diameter lie] outside [the surface, by proposition 14], but on diameter TB one [image lies] on the surface, the rest outside [by proposition 12]. Finally, if arc OG is rotated, it will form a portion [of the sphere] within which some of the images lie on the surface, some outside the sphere, and some inside [by proposition 13]. 
◉ Verum visus non comprehendit que ymagines in superficie spere aut que extra, nec certificatur in comprehensione earum nisi quod sunt ultra portionem apparentem. Iam ergo determinata sunt in hiis speculis ymaginum loca. 
◉ Yet the visual faculty does not perceive which images lie on the surface of the sphere or which [lie] outside, nor does it determine anything in the process of perceiving them except that they lie behind the visible portion [of the sphere].⁑ At this point, then, the imagelocations in these [sorts of] mirrors have been determined. 
◉ [PROPOSITIO 16] Amplius, puncti visi forma non potest in hoc speculo ad visum reflecti nisi a solo puncto speculi. 
◉ [PROPOSITION 16] Furthermore, in this [sort of] mirror, the form of a visible point can be reflected to the eye from only one point on the mirror. 
◉ Sit enim punctus B [FIGURE 5.2.16, p. 571], A centrum visus, et non sit A in perpendiculari ducta ad centrum. Dico quod B refertur ad A ab uno solo puncto speculi, et unam solam pretendit visui ymaginem in hoc speculo. 
◉ Let the [visible] point be B [figure 5.2.16, p. 228], let A be the center of sight, and let A not stand on the normal dropped to the center [of the sphere through the point of reflection]. I say that [the form of] B is reflected to A from only one point on the mirror, and it yields only one image to the eye in this [kind of] mirror. 
◉ Palam quod ab aliquo puncto potest reflecti forma eius. Sit illud G, et ducantur BG, AG. Et sit N centrum spere, et ducatur dyameter BN secans superficiem spere in puncto L, et termini portionis opposite visui sint D, E. Et secet linea AG perpendicularem in puncto Q, qui est locus ymaginis. 
◉ It is clear that its form can be reflected from some point. Let that point be G, and draw [line of incidence] BG and [line of reflection] AG. Let N be the center of the sphere, draw diameter BN [i.e., the normal dropped from objectpoint B] to intersect the sphere’s surface at point L, and let the limits of the [visible] portion [of the mirror] facing the center of sight be D and E [that portion being subtended by tangents AE and AD]. Let line AG intersect normal [BN] at point Q, which is the imagelocation. 
◉ Palam quod A, N, B sunt in eadem superficie ortogonali super speram. Et cum omnes superficies ortogonales super speram in quibus fuerit BN secent se super BN, et non possit superficies in qua linea BN extendi ad punctum A nisi una tantum, palam quod A et B sunt in una superficie ortogonali tantum super speram, non in pluribus. Et cum necesse sit quod punctus visus et A sint in eadem superficie ortogonali super punctum reflexionis, palam quod non fiet reflexio puncti B ad visum nisi in circulo spere qui est in superficie ANB. Sit ergo circulus DGE. Dico iterum quod a nullo puncto huius circuli, preter quam a G, fiet reflexio. [2.129] Si dicatur quod a puncto L, cum BN sit perpendicularis, et AL non sit perpendicularis, et forma per perpendicularem veniens necessario per perpendicularem reflectatur, planum quod non refertur B ad A a puncto L. Iterum nec ab alio puncto arcus LE. Quia ad quodcumque punctum illius arcus ducatur linea a puncto B, tenebit cum contingente illius puncti angulum obtusum ex parte E, et linea ducta a puncto A ad illud punctum tenebit cum contingente illa angulum acutum ex parte L. Quare si ab illo puncto fieret reflexio, esset angulus acutus equalis obtuso. 
◉ It is obvious that A, N, and B lie in the same plane orthogonal to the sphere. And since all planes that are orthogonal to the sphere and that contain BN intersect along BN, and since only one plane containing line BN can be extended through point A, it is clear that A and B lie in only one plane that is orthogonal to the sphere, not in several [such planes]. Moreover, since the visible point [B] and [center of sight] A must lie in the same plane that is orthogonal [to the sphere] at the point of reflection, it is clear that the reflection of [the form of] point B to the eye will occur only on the [great] circle that lies in plane ANB within the sphere. Accordingly, let that circle be DGE. I say, again, that reflection will occur from no point other than G on this circle. 
◉ Iterum a nullo puncto GL potest fieri reflexio. Sumatur enim punctum quodcumque, et sit Z, et ducatur linea AZO secans perpendicularem in puncto O. Et ducatur linea contingens circulum in puncto Z, que cadit necessario inter BG et BL, et sit MZ. Et sit FG linea contingens circulum in puncto G. Palam ex superioribus quod proportio BN ad NQ sicut BF ad FQ. Eodem modo erit proportio BN ad NO sicut proportio BM ad MO. Sed maior est proportio BN ad NQ quam BN ad NO. Igitur maior est proportio BF ad FQ quam BM ad MO, quod plane impossibile, cum BF sit minus BM, et FQ maius MO. Restat ergo ut a puncto Z non fiat reflexio. [2.131] Verum quod ab aliquo puncto arcus GD non fiat reflexio sic constabit. Sumatur quodcumque punctum, et sit T. Educatur linea BT, et linea ATH secans BN in puncto H. Et ducatur contingens circulum in puncto T, que sit CT. Erit ergo proportio BN ad NH sicut BC ad CH, et BN ad NQ sicut BF ad FQ. Sed maior BN ad NH quam BN ad NQ. Ergo maior BC ad CH quam BF ad FQ, quod plane falsum, cum BF maior BC, et CH maior FQ. Restat ergo ut a nullo puncto arcus GD fiat reflexio puncti B, quare quodlibet punctum ab uno solo puncto spere refertur ad visum. Ergo una sola erit linea reflexionis cuiuslibet puncti visi, quare unica unius puncti ymago. 
◉ Likewise, reflection cannot occur from any point on [arc] GL. Take some [such] point, let it be Z, and draw line AZO to intersect normal [BN] at point O. Then draw the line tangent to the circle at point Z, a line that necessarily falls between [i.e., intersects] BG and BL, and let it be MZ. Let FG be the line tangent to the circle at point G. It is clear from the above that BN:NQ = BF:FQ [from proposition 7 above, F being the endpoint of tangency]. So too, BN:NO = BM:MO [from proposition 7 above, Z being the point of reflection and M the endpoint of tangency]. But BN:NQ > BN:NO [by Euclid, V.8]. Therefore BF:FQ > BM:MO, which is clearly impossible, since BF < BM, while FQ > MO. It therefore follows that reflection may not occur from point Z. 
◉ Si autem punctus B fuerit in perpendiculari visuali, palam quod reflectitur ab uno solo puncto, quoniam per perpendicularem tantum, et unica erit eius ymago et propter continuitatem aliorum punctorum in loco ymaginis proprio. 
◉ Moreover, if point B lies on the visual axis [which is the normal dropped from the eye to centerpoint N of the sphere], it is obvious that it is reflected from only one point, because [it is reflected] only along the normal, and its image will be unique and will lie in the appropriate imagelocation because of its continuity with the other points [as established in proposition 8]. 
◉ [PROPOSITIO 17] Amplius, si in aliquo dyametro sumantur duo puncta ex eadem parte centri, locus ymaginis puncti centro propinquioris erit remotior a centro spere loco ymaginis puncti remotioris a centro spere. Et locus reflexionis puncti propinquioris centro erit remotior a centro visus quam puncti remotioris a centro spere. 
◉ [PROPOSITION 17] Furthermore, if two points are taken on a given diameter on the same side of the [sphere’s] center, the imagelocation of the point nearer the [sphere’s] center will lie farther from the center of the sphere than the imagelocation of the point farther from the sphere’s center. Also, the point of reflection for the point nearer the [sphere’s] center will lie farther from the center of sight than the [point of reflection] for the point farther from the center of the sphere. 
◉ Verbi gratia, dico quod locus ymaginis puncti C [FIGURE 5.2.17, p. 572] remotior est a centro loco ymaginis puncti B, et punctus reflexionis puncti C remotior ab A puncto puncto reflexionis puncti B, qui est punctus G. Dico quoniam punctus C non reflectitur nisi ab aliquo puncto arcus GL. 
◉ I say, for example, that the imagelocation for point C [figure 5.2.17, p. 229] lies farther from the center [N] than does the imagelocation for point B, and that the point of reflection for point C lies farther from point A [the center of sight] than the point of reflection for point B, i.e., point G. I say [further] that [the form of] point C is reflected [to A] from a point on arc GL only. 
◉ Palam enim quod non reflectetur ab aliquo puncto arcus LE, nec a puncto L, nec a puncto G, cum B reflectitur ab eo. Et si dicatur quod ab aliquo puncto arcus GD, sit ille punctus T, et sit CT linea per quam forma movetur ad speculum. Et ducatur perpendicularis NT per que necessario dividet angulum CTA per equalia, et ducatur perpendicularis NGK. Erit angulus NTA maior NGA. Restat ergo angulus PTA minor angulo KGA, quare angulus CTP minor angulo BGK. Sed angulus CTP valet angulum TNC et angulum TCN, quia extrinsecus. Et angulus BGK valet angulum GNB et angulum GBN. Erunt ergo duo anguli TNC, TCN minores duobus angulis GBN, GNB, quod est impossibile, cum angulus TNC contineat GNB tanquam partem et angulus TCN sit maior GBN. 
◉ For it is evident [from the previous theorem] that it will not be reflected from any point on arc LE, nor from point L, nor from point G, since [the form of] B is reflected from that point. But if it is claimed that [it is reflected] from some point on arc GD, let that point be T, and let CT be the line along which the form [of C] radiates to the mirror. Then draw normal NT, which will necessarily bisect angle CTA, and draw normal NGK. Angle NTA > [angle] NGA [by Euclid, I.21]. It therefore follows that angle PTA [adjacent to angle NTA] < angle KGA [adjacent to angle NGA], so angle CTP < angle BGK [because, by supposition, angle CTP = angle PTA, and angle BGK = angle KGA, and we have just concluded that angle PTA < angle KGA]. But [by Euclid, I.32] angle CTP = angle TNC + angle TCN, because it is an exterior [angle of triangle TCN]. In addition, angle BGK [which is an exterior angle of triangle BGN] = angle GNB + angle GBN. Accordingly, the [sum of the] two angles TNC and TCN < the [sum of the] two angles GBN and GNB , which is impossible, since angle TNC contains [angle] GNB as a part, while angle TCN > [angle] GBN. 
◉ Restat ergo ut punctus C non reflectatur nisi a punctis G et L intermediis. Et omnes linee a puncto A per hec puncta ducte ad dyametrum BN cadunt in puncta a centro spere remotiora puncto Q, et cadunt in puncta spere a centro visus magis elongata puncto G, et ita propositum. 
◉ Hence, it follows that point C may only be reflected from [some point among the] points lying between G and L. All lines drawn from point A to normal BN through such points fall on points [between L and Q] that lie farther than point Q from the center of the sphere, and they fall to points on the sphere [between L and G] that are farther away from the center of sight [A] than point G, and so what was proposed [has been proven]. 
◉ [PROPOSITIO 18] Amplius, dato speculo et dato puncto viso, est invenire punctum reflexionis. 
◉ [PROPOSITION 18] Now, given a [convex spherical] mirror and given a visible point, to find the point of reflection. 
◉ Sit enim B [FIGURE 5.2.18, p. 572] punctus visus, A centrum visus, et ducantur ab eis due linee ad centrum speculi. Si fuerint ille linee equales, erit facile invenire. Quoniam sumetur circulus spere in superficie illarum linearum, et scimus quod ab uno solo puncto illius circuli fit reflexio. Dividetur ergo angulus quem continent in centro ille due linee per equalia. 
◉ Let B [figure 5.2.18, p. 229] be the visible point, [let] A [be] the center of sight, and draw two lines from these points to the center of the mirror. If those lines are equal, it will be easy to find [the point of reflection]. For a [great] circle on the sphere will be defined in the plane within which those two lines lie, and we know that reflection occurs from only one point on that circle. Accordingly, the angle those two lines form at the [sphere’s] center will be bisected. 
◉ Et ducatur linea dividens angulum extra speram. Erit quidem perpendicularis super lineam contingentem hunc circulum in puncto per quem transit. Et si ducantur ad illud punctum due linee, una a centro visus alia a puncto viso, efficient cum perpendiculari illa et duabus primis lineis duos triangulos, quorum duo latera duobus lateribus equalia, et angulus angulo. Et ita punctus circuli per quem transit illa perpendicularis est punctus reflexionis, quod est propositum. 
◉ Draw the line [GN] that bisects the angle [and extend it] beyond the sphere. It will of course be perpendicular to the line tangent to this circle at the point [G] through which it passes. And if two lines [AG and BG] are drawn to that point, one from the center of sight and the other from the visible point, they will form two triangles with that normal and the two original lines, two sides of these triangles [i.e., BN, NG and AN, NG] being equal [respectively] and angle [BNG of the one, being equal] to angle [ANG of the other].⁑ Hence, the point on the circle through which that normal passes constitutes the point of reflection, which is what was set out [to be demonstrated]. 
◉ Si vero linea a puncto viso ad centrum spere ducta fuerit inequalis linee a centro visus ad idem centrum ducte, oportet nos quedam antecedentia proponere, quorum unum est: 
◉ On the other hand, if the line extending from the visible point to the center of the sphere is not the same length as the line extending from the center of sight to that same centerpoint, we must set forth some preliminary things, one of which is as follows. 
◉ [PROPOSITIO 19] Sumpto circuli dyametro et sumpto in circumferentia puncto, est ducere ab eo ad dyametrum extra productum lineam que a puncto in quo secat circulum usque ad concursum cum dyametro sit equalis linee date. 
◉ [PROPOSITION 19, LEMMA 1] If a diameter is taken in a circle, and if a point is taken on its circumference, a line can be drawn from that point to the extension of the diameter beyond [the circle] such that its extension from the point where it intersects the circle to the point where it intersects the diameter is equal to a given line. 
◉ Verbi gratia, sit QE [FIGURE 5.2.19a, p. 573] data linea, GB dyameter circuli ABG, A punctus datus. Dico quod a puncto A ducam lineam que a puncto in quo secaverit circulum usque ad dyametrum GB sit equalis linee QE, quod sic constabit. Ducantur due linee AB, AG que aut erunt equales aut inequales. 
◉ For instance, let QE [figure 5.2.19, p. 230] be the given line, GB the diameter of circle ABG, and A the given point [on its circumference]. I say that I can draw a line from point A such that [the segment extending] from the point where it intersects the circle to diameter GB is equal to line QE, which will be proven as follows. Draw the two lines AB and AG, which are either equal or unequal. 
◉ Sint equales, et adiungatur linee QE linea talis ut illud quod fiet ex ductu totius cum adiuncta in adiunctam sit equale quadrato AG, et sit linea adiuncta EZ. Cum igitur illud quod fiet ex ductu QZ in EZ sit equale ei quod fit ex ductu AG in se, erit QZ maior AG. Si enim EZ fuerit equalis aut maior AG, est impossibile ut ductum QZ in EZ sit equale quadrato AG. Si autem minor, palam quod QZ maior AG. 
◉ Let them be equal, and add a line to QE—let EZ be this additional line—such that the rectangle that will be formed by the whole line augmented by the additional line and the additional line [i.e., (QE + EZ),EZ = QZ,EZ] = AG2.⁑ Accordingly, since QZ,EZ = AG2, QZ > AG. For if EZ were equal to, or longer than AG, it is impossible that QZ,EZ = AG2 [because the whole, QZ, is greater than its part, EZ, whose square = AG2]. On the other hand, if it is shorter, it is clear that QZ > AG. 
◉ Producatur ergo AG ad equalitatem, et sit AGT. Et posite pede circini super A, fiat circulus secundum quantitatem AGT, qui quidem circulus secabit dyametrum BG, et secet in puncto D. Et ducatur linea AD, que secabit necessario circulum, si enim esset contingens in puncto A, esset equidistans BG, et numquam concurreret cum ea. Secet igitur in puncto H, et ducatur linea GH. 
◉ Accordingly, extend AG [figure 5.2.19a, p. 230] until it is equal [to QZ], and let the result be AGT. Then, placing a compasspoint at A, draw a circle according to radius AGT, and [since] this circle will intersect diameter BG, let it intersect [it] at point D. Then draw line AD, which will necessarily intersect circle [AGB], for if it were tangent at point A, it would be parallel to BG and would never intersect it. Accordingly, let it intersect [the circle] at point H, and draw line GH. 
◉ Palam, cum ABGH sit quadrangulum intra circulum, duo anguli oppositi, scilicet ABG, AHG, valent duos rectos. Sed AGB est equalis ABG, cum respiciant equalia latera, ex ypothesi. Erit igitur angulus AHG equalis angulo DGA, et angulus HAG communis triangulo totali ADG et partiali AHG. Restat ergo ut angulus HDG sit equalis angulo HGA, et triangulus sit similis triangulo, quare proportio DA ad AG sicut AG ad AH. Igitur quod fit ex ductu DA in HA est equale quadrato AG. Sed DA equalis TA; igitur est equalis QZ. Et erit AH equalis EZ et DH equalis QE, que est data linea, et ita propositum. 
◉ It is clear that, since ABGH constitutes a quadrilateral within the circle, the two opposite angles, i.e., ABG and AHG, sum up to two right angles [by Euclid, III.22]. But [angle] AGB = [angle] ABG, since they are subtended by equal sides, according to construction. Therefore, angle AHG = angle DGA [because, as previously concluded, angle AHG = two right angles – angle ABG, and angle DGA = two right angles – angle AGB, which = angle ABG], and angle HAG is common to the whole triangle ADG and triangle AHG that forms part of it. It therefore follows that angle HDG = angle HGA and that [the one] triangle [DGA] is similar to [the other] triangle [AHG], so [by Euclid, VI.4] DA:AG = AG:AH. Hence, DA,HA = AG2 [by Euclid, VI.17]. But DA = TA [since they are both radii of circle TD], so DA = QZ [since TA = QZ, by construction]. And [therefore] AH = EZ, and DH = QE, which is the given line, and so what was proposed [has been demonstrated].⁑ 
◉ Si vero AB et AG non sint equales [FIGURE 5.2.19c, p. 573], protrahatur a puncto G linea equidistans AB que sit GN, et sumatur linea quecumque, que sit ZT, et fiat super punctum Z angulus equalis angulo AGD per lineam ZF. Et ducatur a puncto T linea equidistans ZF, et sit TM, et ex angulo TZF secetur angulus equalis angulo NGD per lineam ZM, quoniam hec linea necessario concurret cum TM, cum sit inter equidistantes. Et sit punctus concursus M. Restat igitur angulus MZF equalis angulo AGN. 
◉ Now, if AB and AG are not equal [figure 5.2.19c, p. 231], draw a line from point G that is parallel to AB, let it be GN, and take some line ZT, and form an angle at point Z with line ZF [i.e., angle TZF] equal to angle AGD. Then, from point T draw a line parallel to ZF, let it be TM, and from angle TZF cut off an angle with line ZM [i.e., angle MZT] equal to angle NGD, for this line [MZ] will necessarily intersect TM, since it lies between parallels [TM and ZF]. Let the point of intersection be M. It follows, therefore, that angle MZF = angle AGN [since, by construction, angle TZF = angle AGD, and angle TZM = angle NGD, so angle MZF (which = angle TZF – angle TZM) = angle AGN (which = angle AGD – angle NGD)]. 
◉ Et a puncto T ducatur linea equidistans linee ZM, que sit TO, que quidem necessario concurret cum FZ. Et sit concursus in puncto K. Et sumatur linea cuius proportio ad lineam ZT sicut BG ad EQ, lineam datam, et sit I. Deinde fiat super punctum M sectio piramidalis quemadmodum docet Ablonius in libro secundo de piramidalibus, propositione quarta, et sit UCM, que quidem sectio non secet lineas KO, KF. Et in hac sectione ducatur linea equalis linee I, scilicet MC, et producatur usque ad lineas KT, KF, et sint puncta sectionum O, L. Igitur, sicut ibidem probabitur, erit OM equalis CL . 
◉ Now, from point T draw a line TO parallel to line ZM, and this line will necessarily intersect FZ. Let the point of intersection be K. Take some line I that is to line ZT as [line] BG is to [line] EQ, [which is the] given line [i.e., I:ZT = BG:EQ, by Euclid, VI.12]. Then, at point M produce a conic section [i.e., a hyperbola] in the way Apollonius describes in the fourth proposition of book 2 of his Conics, and let this conic section be UCM, which may not intersect lines KO and KF [that form its asymptotes].⁑ In this section draw a line equal to line I, i.e., MC, extend it to lines KT and KF, and let O and L be the intersectionpoints [where MC extended cuts asymptotes KT and KF]. Thus, as the same [book of Apollonius’ Conics] will demonstrate, OM = CL.⁑ 
◉ Et a puncto T ducatur linea equidistans CM, que sit TF, et super punctum A fiat angulus equalis angulo ZFT per lineam AND. Palam quod hec linea concurret cum GD, cum angulus AGN sit equalis FZM angulo, et angulus GAN angulo ZFT. Igitur AD linea aut erit contingens circulo aut secabit ipsum, quoniam, si non fuerit contingens, et arcus AB fuerit maior arcu AG, secabit arcum AB, et si AB fuerit minor, secabit arcum AG. 
◉ From point T draw line TF parallel to [line] CM, and at point A form an angle [GAN] with line AND that is equal to angle ZFT. It is evident that this line [AND] will intersect GD, since angle AGN = angle FZM, and angle GAN = angle ZFT. Hence, line AD will either be tangent to the circle or will intersect it, because, if it is not tangent, and if arc AB > arc AG, it will cut arc AB, whereas if AB < [AG], it will cut arc AG. 
◉ Sit igitur contingens in puncto A. Cum angulus GAN sit equalis angulo ZFT, et angulus AGN angulo FZY, erit tertius tertio equalis, et erit triangulus AGN similis triangulo ZFY. Similiter, cum AGD sit equalis angulo FZT, erit triangulus AGD similis triangulo FZT. Igitur que est proportio AN ad AG ea est proportio FY ad FZ, et que est proportio AG ad GD ea FZ ad ZT, quare que est proportio AN ad GD ea est FY ad ZT. 
◉ Accordingly, let it be tangent at point A. Since angle GAN [in triangle GAN] = angle ZFT [in triangle ZFY], and since angle AGN = angle FZY, the third angle [ANG] = the third angle [ZYF], so [by Euclid, VI.4] triangle AGN will be similar to triangle ZFY. Likewise [in triangles AGD and FZT], since [angle] AGD = angle FZT [while angle ZFT = angle GAD, and the remaining angles ZTF and ADG are equal], then triangle AGD will be similar to triangle FZT. Therefore AN:AG = FY:FZ, and AG:GD = FZ:ZT, so AN:GD = FY:ZT [by Euclid, V.22]. 
◉ Verum, cum TM sit equidistans FL, et FT equidistans ML, erit FT equalis ML, quare erit equalis CO, cum sit MO equalis LC. Sed MO est equalis YT, cum sit ei equidistans, et YM equidistans TO. Restat ergo FY equalis CM. Sed CM equalis I. Erit igitur FY equalis I. Sed proportio I ad ZT sicut BG ad EQ. Igitur proportio AN ad GD sicut BG ad EQ. 
◉ But since TM is parallel to FL [by construction], and since FT is parallel to ML [by construction], FT = ML, so it will be equal to CO, since MO = LC [by Apollonius, II.8]. But MO = YT, since it is parallel to it, and YM is parallel to TO. It therefore follows that FY = CM. But CM = I [by construction]. Thus, FY = I. But I:ZT = BG:EQ [by construction]. Therefore, AN:GD = BG:EQ [because we established earlier that AN:GD = FY:ZT]. 
◉ Verum angulus GAN est equalis angulo GBA, sicut probat Euclides in tertio. Sed angulus NGD est equalis angulo ABG, cum NG sit equidistans AB. Igitur angulus NGD equalis est angulo NAG, et angulus NDG communis, quare tertius tertio equalis, quare triangulus NDG similis triangulo ADG. Igitur proportio AD ad GD sicut GD ad ND, quare quod fit ex ductu AD in DN est equale quadrangulo DG. 
◉ However, angle GAN = angle GBA, as Euclid demonstrates in the third [book of the Elements, prop. 32]. But angle NGD = [alternate] angle ABG, since NG is parallel to AB [by construction]. Therefore, angle NGD [in triangle NDG] = angle NAG [in triangle AGD], and angle NDG is common [to both triangles], so the third [angle, i.e., DNG] = the third [angle, i.e., AGD], and so triangle NDG is similar to triangle ADG. Hence, AD:GD = GD:ND, so AD,DN = DG2 [by Euclid, VI.17]. 
◉ Verum quadratum AD est equale ei quod fit ex ductu BD in DG, sicut probat Euclides, et quadratum AD est equale ei quod fit ex ductu AD in DN et ei quod fit ex ductu AD in NA. Et illud quod fit ex ductu BD in DG est equale quadrato DG et ei quod fit ex ductu BG in GD, sicut probat Euclides. Ablatis ergo equalibus, restat ut quod fit ex ductu AD in AN sit equale ei quod fit ex ductu BG in DG. Igitur proportio secundi ad quartum sicut tertii ad primum, quare proportio AN ad DG sicut BG ad AD. Sed iam dictum est quod proportio AN ad GD sicut BG ad EQ. Igitur EQ equalis AD, quod est propositum. 
◉ But AD2 = BD,DG, as Euclid demonstrates [in III.36], and AD2 = AD,DN + AD,NA [by Euclid, II.2]. And BD,DG = DG2 + BG,GD, as Euclid demonstrates [in II.3]. Therefore, when equal terms are subtracted [i.e., GD2], it follows that AD,AN = BG,DG.⁑ Hence [by Euclid, VI.16], the second [AN] is to the fourth [DG] as the third [BG] is to the first [AD], so AN:DG = BG:AD. But it has already been established that AN:GD = BG:EQ. Therefore, EQ = AD, which is what was proposed. 
◉ Quod si AD non fuerit contingens circulum sed secans, et fuerit AG maior AB [FIGURE 5.2.19d, p. 574], secabit quidem AG. Secet ergo in puncto H, et ducatur linea AG. 
◉ If, however, AD is not tangent to the circle but cuts it, and if AG > AB [figure 5.2.19d, p. 232], then it will cut [arc] AG. Let it cut at point H, and draw line AG. 
◉ Palam quod duo anguli AHG, ABG valent duos rectos. Sed angulus NGD est equalis angulo ABG. Igitur angulus AHG et angulus NGD sunt equales duobus rectis. Quare angulus NGD est equalis angulo NHG, et angulus NDG communis, quare tertius angulus tertio angulo equalis est, et triangulus HGD similis triangulo NDG. Igitur proportio HD ad DG sicut proportio DG ad DN, quare illud quod fit ex ductu HD in DN est equale quadrato GD. 
◉ It is obvious that the two angles AHG and ABG sum up to two right angles [by Euclid, III.22]. But angle NGD = [alternate] angle ABG [because NG is parallel to AB, by construction]. Thus, angle AHG + angle NGD = two right angles. Accordingly, angle NGD = angle NHG [since angle AHG + angle NHG = two right angles], and angle NDG is common [to triangles NGD and HGD], so the third angle [DNG] = the third angle [DGH], and so triangle HGD is similar to triangle NDG. Therefore, HD:DG = DG:DN, so HD,DN = GD2 [by Euclid, VI.17]. 
◉ Sed quod fit ex ductu AD in DH est equale ei quod fit ex ductu BD in DG, sicut probat Euclides, et illud quod fit ex ductu AD in DH est equale ei quod fit ex ductu DH in DN et DH in AN. Et quod fit ex ductu BD in DG est equale ei quod fit ex ductu BG in GD et quadrato GD. Ablatis igitur equalibus (scilicet quadrato GD et eo quod fit ex ductu DH in DN) restat ut illud quod fit ex ductu DH in AN est equale ei quod fit ex ductu BG in DG, quare proportio secundi ad quartum (id est AN ad GD) sicut tertii ad primum (id est BG ad DH). Sed iam probatum est quod proportio AN ad DG sicut BG ad EQ. Igitur EQ est equalis DH, et ita propositum. 
◉ But AD,DH = BD,DG, as Euclid demonstrates,⁑ and AD,DH = DH,DN + DH,AN [by Euclid, II.1]. Moreover, BD,DG = BG,GD + GD2 [by Euclid, II.3]. Therefore, when equal terms are subtracted (i.e., GD2 and DH,DN), it follows that DH,AN = BG,DG,⁑ so [by Euclid, VI.16] the second term is to the fourth (i.e., AN:GD) as the third is to the first (i.e., BG:DH). But it has already been proven that AN:DG = BG:EQ. Thus, EQ = DH, and so what was proposed [has been demonstrated]. 
◉ Si vero AG erit minus AB (et secet ad arcum AB), sit sectio punctus H [FIGURE 5.2.19e, p. 574], et ducatur linea HG. Palam quod angulus NGD est equalis angulo ABG. Sed anguli ABG, AHG sunt equales, quia cadunt in eundem arcum. Igitur angulus NGD equalis est angulo AHG, et angulus NDG communis, quare tertius tertio equalis, et trianguli similes. Igitur proportio HD ad GD sicut GD ad DN, quare quod fit ex ductu HD in DN est equale quadrato GD. 
◉ On the other hand, if AG < AB (and let [AD] cut [the circle] on arc AB), then let H [figure 5.2.19e, p. 232] be the point of intersection, and draw line HG. It is evident that angle NGD = angle ABG [by construction]. But [by Euclid, III.21] angles ABG and AHG are equal, since they are subtended by the same arc [AG]. Therefore, angle NGD = angle AHG, and angle NDG is common [to triangles NGD and HGD], so the third angle [GND] = the third angle [HGD], and the triangles [NGD and HGD] are similar. Accordingly HD:GD = GD:DN, so HD,DN = GD2 [by Euclid, VI.17]. 
◉ Sed quod fit ex ductu HD in DA est equale ei quod fit ex ductu BD in DG, et quod fit ex ductu HD in DA est equale ei quod fit ex DN in HD et AN in HD. Et ductus BD in DG valet quadratum DG et ductum BG in DG. Igitur, remotis equalibus, erit ductus HD in NA sicut BG in DG. Igitur proportio AN ad DG sicut BG ad HD. Sed iam dictum est quod proportio AN ad DG est sicut BG ad EQ. Igitur EQ est equalis HD, quod est propositum, quia a puncto A dato duximus lineam secantem circulum, et a puncto sectionis ad dyametrum est equalis linee date. 
◉ But HD,DA = BD,DG [since, by Euclid, III.36, both are equal to the square on the tangent drawn from D], and HD,DA = DN,HD + AN,HD [by Euclid, II.1]. Moreover BD,DG = DG2 + BG,DG [by Euclid, II.3]. Hence, with equal terms subtracted [i.e., GD2 and HD,DN], HD,NA = BG,DG. Accordingly, AN:DG = BG:HD [by Euclid, VI.16]. But it has already been established that AN:DG = BG:EQ. Therefore, EQ = HD, which is what was proposed, because from point A we have drawn a line to cut the circle, and [the segment] from the point of intersection [H] to the diameter [BG] is equal to the given line. 
◉ [PROPOSITIO 20] Amplius, a puncto dato in circulo extra dyametrum eius est ducere lineam per dyametrum ad circulum, ut pars eius a dyametro ad circulum sit equalis linee date. 
◉ [PROPOSITION 20, LEMMA 2] Now, from a given point on a circle outside its diameter, a line can be drawn through the diameter to the circle so that the segment on it between the diameter and the circle is equal to a given line. 
◉ Verbi gratia ABG [FIGURE 5.2.20, p. 575] sit circulus datus, BG dyameter, A punctus datus, HZ linea data. Dico quod a puncto A est ducere lineam transeuntem per dyametrum BG cuius pars a dyametro ad circulum sit equalis linee HZ. 
◉ For instance, let ABG [figure 5.2.20, p. 233] be the given circle, BG its diameter, A the given point, and HZ the given line. I say that a line can be drawn from point A to pass through diameter BG such that the segment [ED] from the diameter to the circle is equal to line HZ. 
◉ Probatio: ducantur linee AB, AG, et super punctum H fiat angulus equalis angulo AGB per lineam MH, et super idem punctum fiat angulus equalis angulo ABG per lineam HL. Et a puncto Z ducatur equidistans linee HM, que sit ZN, que quidem secabit HL, et a puncto Z ducatur linea equidistans HL, que sit ZT, et secet HM in puncto T. Et a puncto T ducatur sectio piramidalis TP, quam assignabit Ablonius in libro piramidis, que quidem sectio non continget aliquam linearum ZN, HL inter quas iacet. Similiter fiat sectio piramidalis ei opposita inter easdem lineas, que sit CU. 
◉ The proof [is as follows]. Draw lines AB and AG, and at point H form with line HM an angle [MHZ] equal to angle AGB, and at the same point form with line HL an angle [LHZ] equal to angle ABG. Then, from point Z draw a line ZN parallel to line HM, and it will intersect HL; and from point Z draw line ZT parallel to HL, and let it intersect HM at point T. At point T construct conic section [i.e., hyperbola] TP, which Apollonius will describe in his book on conics [II.4], and that [conic] section will never touch either of the lines ZN and HL [i.e., the asymptotes] between which it lies. Between these same lines construct conic section [i.e., hyperbola] CU facing the first one. 
◉ Cum igitur linea minima ex lineis a puncto T ad sectionem CU ductis fuerit equalis dyametro BG, circulus factus secundum hanc minimam lineam, posito pede circini super punctum T, continget sectionem CU. Si vero minima ex lineis a puncto T ad sectionem CU ductis fuerit minor dyametro BG, circulus factus modo predicto secundum quantitatem BG secabit sectionem CU in duobus punctis. 
◉ Thus, if the shortest of all the lines extending from point T to [conic] section CU is equal to diameter BG, a circle produced according to this shortest line [as radius] when the point of a compass is placed at point T will be tangent to [conic] section CU. On the other hand, if the shortest of all the lines extending from point T to [conic] section CU is shorter than diameter BG, the circle produced in the aforementioned way according to [radius] BG will intersect [conic] section CU at two points.⁑ 
◉ Sit ergo CT minima et equalis dyametro BG, que quidem secabit ZQ et HF, cum ducatur ad sectionem que eas interiacet. Et ducatur a puncto Z equidistans huic que quidem secabit HM et HL sicut sua equidistans. Secet ergo in punctis M, L, et sit MZL, et punctus sectionis in quo CT secat ZN sit Q, et super dyametrum GB fiat angulus equalis angulo HLZ, qui sit DGB. Et ducantur due linee AD, BD. 
◉ Accordingly, let CT be the shortest [line], and [let it be] equal to diameter BG, and it will intersect ZQ and HF when it is extended to the [conic] section lying between them.⁑ From point Z draw a line parallel to this one, and it will intersect HM and HL just as its parallel [CT] does. Let it be MZL, let it intersect [those lines] at points M and L, let Q be the point where CT intersects ZN, and on diameter GB form an angle equal to angle HLZ, and let it be DGB. Then draw the two lines AD and BD. 
◉ Palam cum angulus GAB sit rectus, alii duo anguli trianguli AGB valent rectum, quare angulus LHM est rectus, et est equalis angulo GDB. Et angulus HLM est equalis angulo DGB. Igitur tertius tertio, et triangulus similis triangulo, quare proportio GB ad BD sicut LM ad MH. 
◉ It is evident that, since angle GAB is a right angle [by Euclid, III.31], the other two angles in triangle AGB sum up to a right angle, so angle LHM is right [since it consists of the two angles MHZ and ZHL that were constructed equal to angles AGB and ABG, which sum up to a right angle], and it is equal to angle GDB [which is a right angle by Euclid, III.31]. Also [by construction], angle HLM [in triangle HLM] = angle DGB [in triangle DGB]. Hence, the third [angle HML is equal] to the third [angle DBG], and the [first] triangle [HML] is similar to the [second] triangle [GDB], so GB:BD = LM:MH. 
◉ Sed quoniam angulus ADB equalis est angulo BGA, quia cadunt in eundem arcum, et angulus BGA equalis angulo MHZ, ex ypothesi, erit angulus ADB equalis angulo MHZ. Et iam habemus quod angulus GBD est equalis angulo HMZ. Ergo tertius tertio, et triangulus DEB similis triangulo MHZ. Sit E punctus in quo linea AD secat dyametrum BG. Igitur proportio BD ad DE sicut MH ad HZ. Igitur proportio BG ad DE sicut LM ad HZ. 
◉ But, since angle ADB = angle BGA, because they are subtended by the same arc [AB], and since angle BGA = angle MHZ, by construction, then angle ADB = angle MHZ. In addition, we already know that angle GBD = angle HMZ [by the similarity of triangles HML and DGB]. Hence, the third [angle in triangle DEB, i.e., angle DEB, is equal] to the third angle [in triangle MHZ, i.e., angle MZH], so triangle DEB is similar to triangle MHZ. Let E be the point where line AD intersects diameter BG. Accordingly, BD:DE = MH:HZ. Hence, BG:DE = LM:HZ.⁑ 
◉ Verum Ablonius probat quod, cum fuerint due sectiones opposite piramidales inter duas lineas, et producetur linea ab una sectione ad aliam, pars eius que interiacet unam sectionem et unam ex lineis est equalis alii parti que interiacet aliam sectionem et aliam lineam, quare QC equalis TF. Sed TQ est equalis MZ, cum sit ei equidistans et inter duas equidistantes. Igitur MZ equalis FC, et ZL equalis TF. Igitur ML equalis TC, quare proportio GB ad ED sicut TC ad HZ, et cum TC sit equale BG, erit ED equalis HZ, quod est propositum. 
◉ However, Apollonius demonstrates [in Conics, II.16] that, when two [hyperbolic] conic sections lie opposite one another between two [asymptotic] lines, and when a line is drawn from one section to the other, the segment of that line lying between one of the sections and one of the [asymptotic] lines is equal to the opposite segment lying between the other section and the other [asymptotic] line, so QC = TF. But TQ = MZ, since it is parallel to it [by construction] and lies between two parallels [HM and ZN, which are parallel by construction]. Therefore MZ = FC [since FC = TQ], and ZL = TF [since they are parallel and lie between TZ and HL, which are parallel by construction]. Hence, ML = TC, so GB:ED = TC:HZ, and since TC = BG, ED = HZ, which is what was proposed. 
◉ Si autem linea a T ad sectionem CU ducta et minima fuerit minor dyametro BG, producatur ultra sectionem donec sit equalis. Et secundum quantitatem eius fiat circulus qui quidem secabit sectionem in duobus punctis a quibus linee ducte ad T erunt equales BG. Et a puncto Z ducatur equidistans utrique, et tunc erit ducere a puncto A, modo predicto, duas lineas equales linee date. Erit idem penitus probandi modus. 
◉ However, if the line extending from T to [conic] section CU is the shortest line and is shorter than diameter BG, then extend it beyond the [conic] section until it is equal. Then, according to its length [as radius], produce a circle that will intersect the [conic] section at two points, and from those points the lines extending to T will be equal to BG. Then, from point Z draw a line parallel to both, and in that case two lines equal to the given lines will be drawn from point A in the prescribed manner, and this will be proven in precisely the same way.⁑ 
◉ [PROPOSITIO 21] Amplius, dato triangulo ortogonio, et dato puncto in uno laterum angulum rectum continentium, est ducere a puncto illo lineam ad aliud latus continentium rectum lineam secantem tertium oppositum recto, ita quod pars huius linee interiacens punctum sectionis et latus in quo non est punctus datus se habeat ad partem lateris oppositi recto que est a sectione ad latus in quo est punctus datus sicut data linea ad datam lineam. 
◉ [PROPOSITION 21, LEMMA 3] Moreover, given a right triangle, and given some point on one of the sides forming the right angle, a line can be drawn from that point to the other side forming the right [angle] and intersecting the third side facing the right angle in such a way that the segment of this line that lies between the point of intersection and the side on which the given point does not lie is to the segment of the side opposite the right angle from the [point of] intersection to the side containing the given point as a given line is to a given line. 
◉ Verbi gratia ABG [FIGURE 5.2.21, p. 576] est triangulus datus, cuius angulus ABG rectus, et in latere GB est punctus datus D extra triangulum aut intra. Dico quod a puncto D est ducere lineam secantem latus AG et concurrentem cum latere AB ita quod pars eius interiacens latera AB AG sit eiusdem proportionis ad partem lateris AG que est ab illa linea usque ad punctum G sicut se habet E ad Z, que sunt date linee. 
◉ For instance, ABG [figure 5.2.21, p. 235] is the given triangle with ABG the right angle, and the given point D, which is on side GB, is either in or outside the triangle. I say that from point D a line can be drawn to cut side AG and intersect side AB in such a way that the segment [TQ] on it between sides AB and AG is to the segment [TG] of the side AG extending from that line to point G in the same proportion as E is to Z, those being the given lines [i.e., TQ:TG = E:Z]. 
◉ Probatio: sit punctus D in ipso triangulo ABG, et ducatur ab eo linea equidistans AB, que sit DM. Et fiat circulus super tria puncta G, M, D, et protrahatur linea AD. Et quoniam planum quod angulus GMD est equalis angulo GAB, erit maior angulo GAD. Secetur ex eo equalis per lineam MN, et sit DMN, et fit H linea ad quam se habeat AD sicut se habet E ad Z. Et a puncto N, qui est punctus circuli, ducatur linea ad dyametrum GM equalis linee H secundum supradicta, et sit NL, et punctus in quo secat circulum sit C. Et ducatur linea GC, et a puncto D ducatur linea ad punctum C, que, cum cadat inter duas equidistantes, tenens angulum acutum cum altera, si producatur, necessario concurret cum alia. Concurrat igitur, et sit punctus concursus Q. 
◉ The proof [is as follows]. Let point D lie on triangle ABG itself [as represented in figure 5.2.21], and draw from it a line DM that is parallel to AB. Produce a circle upon the three points G, M, and D [by Euclid, IV.5], and draw line AD. Since it is evident that angle GMD = [alternate] angle GAB, it will be greater than angle GAD. With line MN cut from [angle] GMD an angle equal [to GAD], and let it be DMN, and [by Euclid, VI.12] draw line H so that AD is to it as E is to Z [i.e., AD:H = E:Z]. Then, from point N, which lies on the circle, draw a line to diameter GM⁑ that is equal to line H, according to the previous account [in proposition 19, lemma 1], let it be [on line] NL, and let point C be where it intersects the circle [so that LC = H].⁑ Draw line GC, and from point D draw a line to point C, a line which, when extended, necessarily intersects the other line [i.e., AB], since it falls between two parallels [AB and DM, which are parallel by construction] and forms with one of them an acute angle [i.e., MDT]. Let them intersect, then, and let Q be the point of intersection. 
◉ Palam quod angulus GMD est equalis angulo GCD, quia cadunt in eundem arcum, et angulus GMD est equalis angulo GAB. Restat igitur ut angulus GCQ sit equalis angulo GAQ. Sit T punctus in quo DQ secat AG, et angulus GTC est equalis angulo ATQ. Igitur tertius tertio, quare triangulus ATQ similis est triangulo TCG. Igitur proportio QT ad TG sicut AT ad TC. 
◉ It is evident that angle GMD = angle GCD, because they are subtended by the same arc [GND], and angle GMD = angle GAB [by construction]. It therefore follows that angle GCQ = angle GAQ [because they are adjacent to equal angles, i.e., GCD and GAB]. Let T be the point where DQ intersects AG, and [so] angle GTC [in triangle TCG] = [vertical] angle ATQ [in triangle ATQ, while angle GCT (i.e., GCQ) = angle TAQ (i.e., GAQ), as established above]. Hence, the third [angle TGC] = the third [angle TQA], so triangle ATQ is similar to triangle TCG. Thus, QT:TG = AT:TC. 
◉ Verum angulus NMD est equalis angulo TAD et angulo NCD, quare NCD equalis TAD. Et angulus CTL communis duobus triangulis, quare tertius tertio, et triangulus similis triangulo, scilicet TLC triangulo TAD. Igitur proportio TA ad CT sicut proportio AD ad LC, quare erit proportio AD ad LC sicut QT ad TG. Sed LC est equalis linee H, et proportio AD ad H sicut E ad Z. Igitur proportio QT ad TG sicut E ad Z, quod est propositum. 
◉ But angle NMD = angle TAD [i.e., GAD, by construction], and [it is also equal to] angle NCD [because they are both subtended by arc ND], so [angle] NCD [= vertical angle TCL in triangle TCL, and it also] equals [angle] TAD [in triangle TAD]. In addition, angle CTL is common to both triangles, so the third [angle TLC] = the third [angle TDA], and [so] the [one] triangle is similar to the [other] triangle, i.e., [triangle] TLC to triangle TAD. Hence, TA:CT = AD:LC [but TA:CT = QT:TG, as established above], so AD:LC = QT:TG. But LC = line H, and AD:H = E:Z. Therefore, QT:TG = E:Z, which is what was proposed. 
◉ Si vero D sumatur in illo latere extra triangulum [FIGURES 5.2.21a, 5.2.21b, pp. 576 and 577], producatur a puncto D equidistans AB, et sit DM, et ducatur AG donec concurrat cum DM in puncto M. Et fiat circulus transiens per tria puncta G, D, M, et ducatur linea AD. Erit quidem angulus GAD maior angulo GMD. Fiat ei equalis, et sit NMD, et a puncto N, qui sit punctus circuli, ducatur linea equalis H linee ad quam H se habeat AD sicut E ad Z, et sit NCL, et hoc super dyametrum MG. Et concursus sit L. 
◉ If, on the other hand, D is taken on that [same] side [that includes the right angle but extends] beyond the triangle [figures 5.2.21a and 5.2.21b, p. 235], then [in figure 5.2.21a] draw [a line] from point D that is parallel to AB, and let it be DM, and extend AG until it intersects DM at point M. Then produce a circle passing through the three points G, D, and M, and draw line AD. Angle GAD [which is exterior to triangle DAM] will of course be greater than [interior] angle GMD. Form an angle equal to it [i.e., equal to GAD], and let it be NMD, and from point N, which is a point on the circle, draw a line equal to line H such that AD is to that line as E is to Z, and let that line be [on line] NCL, which is extended to diameter MG [so that CL = H and, therefore, that AD:CL = E:Z]. Let the point of intersection be L. 
◉ Cum igitur angulus NMD et angulus NCD valeant duos rectos, et angulus NMD sit equalis angulo TAD, erunt duo trianguli TCL, TAD similes. Et cum duo anguli GCD, GMD sint equales, erunt duo trianguli GCT, TAQ similes, et erit proportio AD ad CL, que est equalis H, sicut QT ad TG, et ita E ad Z sicut QT ad TG, quod est propositum. 
◉ Thus, since angle NMD and angle NCD sum up to two right angles [by Euclid, III.22], and since angle NMD = angle TAD, the two triangles TCL and TAD will be similar.⁑ In addition, since the two angles GCD and GMD are equal [by Euclid, III.21], the two triangles GCT and TAQ will be similar [so that DT:LT = TQ:TG = AD:CL], and so AD:CL (which = H) = QT:TG, and so E:Z = QT:TG, which is what was proposed.⁑ 
◉ [PROPOSITIO 22] Amplius, duobus punctis datis, scilicet E, D, et dato circulo, est invenire punctum in eo ut angulum contentum a lineis a punctis predictis ad illud punctum ductis dividat per equalia linea circulum contingens in puncto illo. 
◉ [PROPOSITION 22, LEMMA 4] Now, given two points, i.e., E and D, and given a circle, to find a point on it such that the line tangent to the circle at that point bisects the angle formed by the lines drawn from the aforementioned points to that point. 
◉ Verbi gratia ducatur a puncto E [FIGURE 5.2.22, p. 578] ad centrum circuli dati linea EG, et producatur usque ad circumferentiam, et sit ES. Deinde ducatur linea GD et sit MI linea in puncto C divisa ut proportio IC ad CM sicut EG ad GD. Et dividatur MI per equalia in puncto N, et ducatur perpendicularis NO. Et super punctum M fiat angulus equalis medietati anguli DGS per lineam MO. Palam quod erit minor recto, et angulus ONM rectus. Igitur MO concurret cum NO. Concurrat autem in O puncto, et a puncto C ducatur linea ad triangulum que sit CKF ita ut proportio KF ad FM sit sicut proportio EG ad GS. Et super punctum G fiat angulus equalis angulo MFK per lineam usque ad circulum ductam, que sit AG, et sit angulus AGE. Et ducantur due linee AG, DG. Dico quod A est punctus quem querimus. 
◉ For example, from point E [figure 5.2.22, p. 236] draw line EG to the center of the given circle, extend it to the circumference, and let [this extended line] be ES. Then, draw line GD, and let line MI be cut at point C so that IC:CM = EG:GD [by Euclid, VI.10]. Then bisect MI at point N, and draw perpendicular NO. On point M form with line MO an angle [OMN] that is half of angle DGS. It is clear that it will be smaller than a right angle, whereas angle ONM [will be] a right angle [by construction]. Thus, MO will intersect NO. Let it intersect at point O, and from point C draw a line CKF to the triangle [NMO] such that KF:FM = EG:GS [by proposition 21, lemma 3, case 2]. Then, with line AG, which is extended to the circle, form an angle at point G that is equal to angle MFK, and let it be angle AGE. Finally, draw the two lines AG and DG. I say that A is the point we are seeking. 
◉ Ducatur linea EA. Cum ergo MFK sit equalis angulo AGE, et proportio KF ad FM sicut proportio EG ad GA, cum GA sit equalis GS, erit triangulus AGE similis triangulo MFK. Igitur, angulus FMK est equalis angulo EAG, et angulus AEG equalis angulo MKF. 
◉ Draw line EA. Accordingly, since [angle] MFK = angle AGE [by construction], and since KF:FM = EG:GA [by construction], because GA = GS [since G is the circle’s center, by construction], then triangle AGE will be similar to triangle MFK [by Euclid, VI.4]. Hence, angle FMK = angle EAG, and angle AEG = angle MKF. 
◉ Igitur a puncto A ducatur linea tenens cum linea AE angulum equalem angulo NMK, et sit linea AZ, que necessario concurret cum linea GE, quoniam que est proportio KF ad FM ea est EG ad GA, et angulus GAZ equalis angulo FMC. Igitur, sicut linea MO concurret cum FK in puncto F, concurret AZ cum GE. Sit concursus in puncto Z, et producatur AZ usque ad punctum Q ita ut linea AZ se habeat ad ZQ sicut MC ad CI, et ducatur linea EQ. 
◉ From point A, then, draw a line that forms with line AE an angle [EAZ] equal to angle NMK, and let it be line AZ, which will necessarily intersect line GE, because KF:FM = EG:GA, and angle GAZ = angle FMC [since angle EAG = angle FMK, and angle EAZ = angle KMN, by construction, so EAG + EAZ = GAZ = FMK + KMC = FMC]. Therefore, just as line MO will intersect [line] FK at point F, AZ will intersect GE. Let the intersection occur at point Z, and draw AZ to point Q so that AZ:ZQ = MC:CI [by Euclid, VI.12], and draw line EQ. 
◉ Deinde a puncto A ducatur equidistans EQ, que sit AT. Erit quidem angulus AQE equalis angulo QAT, et quoniam duo anguli ZEA, EAT sunt minores duobus rectis, concurret AT necessario cum EZ. Sit concursus punctus T. Palam quod angulus AEG est equalis angulo MKF. Ducta a puncto E linea perpendiculari super AZ, que sit EL, erit angulus AEL equalis angulo MKN, cum angulus EAL sit equalis angulo KMN, et angulus ALE equalis angulo MNK, quia uterque rectus. Restat ergo ut angulus LEZ sit equalis angulo NKC, et angulus ELZ rectus equalis angulo KNC. Restat ut angulus EZL sit equalis angulo KCN. Igitur angulus EZQ equalis angulo KCI. 
◉ Then, from point A draw a [line] parallel to EQ, and let it be AT. Angle AQE = [alternate] angle QAT [because QA intersects parallels AT and EQ], and since the two angles ZEA and EAT sum up to less than two right angles [because angle ZEA + adjacent angle AEG = 2 right angles, and EAT < AEG], AT will necessarily intersect EZ. Let T be the point of intersection. It is clear that angle AEG = angle MKF [because, as we concluded above, triangles AGE and MFK are similar]. When line EL is drawn from point E perpendicular to AZ, angle AEL = angle MKN, since [by construction] angle EAL [in triangle EAL] = angle KMN [in triangle KMN], and angle ALE = angle MNK, since both are right angles [so the remaining angles AEL and MKN are equal]. It follows, therefore, that angle LEZ = angle NKC [since both are adjacent to equal angles, i.e., LEG (= AEL + AEG) and NKF (= MKF + MKN), respectively], and right angle ELZ = [right] angle KNC [so triangles ELZ and NKC are similar]. It follows that angle EZL = angle KCN. Therefore, [exterior] angle EZQ [of triangle EZL] = [exterior] angle KCI [of triangle NKC]. 
◉ Palam ergo quod triangulus EAG similis triangulo FMK, et triangulus EAL similis triangulo KMN, et triangulus ELZ similis triangulo KNC, et triangulus EAZ triangulo KMC. Ergo proportio AZ ad ZE sicut MC ad CK, et proportio QZ ad ZA sicut IC ad CM, et proportio QZ ad ZE sicut IC ad CK, quare triangulus QZE similis triangulo ICK, et triangulus QLE similis triangulo IKN. Erit proportio NM ad NI sicut AL ad LQ, et ita AL equalis LQ, et EQ erit equalis EA, et angulus EQZ equalis angulo LAT, et angulus EZQ equalis angulo AZT. Igitur tertius tertio equalis, et triangulus EZQ similis triangulo ZAT, quare proportio QZ ad ZA sicut EZ ad ZT, et sicut EQ ad AT, et sicut AE ad AT. Sed QZ ad ZA sicut EG ad GD. Igitur AE ad AT sicut EG ad GD. 
◉ So it is evident that triangle EAG is similar to triangle FMK, triangle EAL is similar to triangle KMN, triangle ELZ is similar to triangle KNC, and triangle EAZ is similar to triangle KMC. Accordingly, AZ:ZE = MC:CK, and QZ:ZA = IC:CM [by construction], and [so, by Euclid, V.22] QZ:ZE = IC:CK, so triangle QZE is similar to triangle ICK, while triangle QLE is similar to triangle IKN. [Accordingly] NM:NI = AL:LQ [since triangles KNM and AEL are similar], and so AL = LQ [since NM = NI, by construction], EQ = EA, angle EQZ [in triangle EQZ] = angle LAT [in triangle ZAT, by construction], and angle EZQ = [vertical] angle AZT. Hence, the third [angle, i.e., ZEQ] = the third [angle, i.e., ATZ], and triangle EZQ is similar to triangle ZAT, so QZ:ZA = EZ:ZT = EQ:AT = AE:AT [since AE and EQ are equal by previous conclusions]. But QZ:ZA = EG:GD [because EG:GD = IC:CM, by construction, and QZ:ZA = IC:CM, by construction]. Therefore AE:AT = EG:GD. 
◉ Fiat autem supra punctum A angulus equalis angulo GAE, qui sit UAG. Palam quod angulus GAL est medietas anguli UAT, sed est medietas anguli DGU, quare angulus UAT est equalis angulo DGU. Sed anguli TAU, TUA sunt minores duobus rectis, cum AT et UT concurrant, quare duo anguli TUA, DGU sunt minores duobus rectis. Igitur AU concurret cum DG. 
◉ Now, at point A form an angle equal to angle GAE, and let it be UAG. It is clear that angle GAL is half of angle UAT,⁑ but it is [also] half of angle DGU [since, by construction, OMN is half DGU, and GAL = OMN], so angle UAT = angle DGU. But angles TAU and TUA sum up to less than two right angles, since AT and UT intersect, so the two angles TUA and DGU sum up to less than two right angles. Therefore, AU will intersect DG. 
◉ Dico quod concurret in puncto D, quoniam efficiet cum lineis UG, GD triangulum similem triangulo AUT, habebunt enim angulum AUG communem, et angulus TAU equalis angulo UGD. Igitur proportio AU ad AT sicut UG ad lineam quam secat AU ex GD, et proportio EA ad AU sicut EG ad GU, cum sit angulus UAG equalis angulo GAE. 
◉ I say that it will intersect at point D, because [by Euclid, VI.4] it will form a triangle with lines UG and GD that is similar to triangle AUT, for they will have angle AUG in common, and angle TAU = angle UGD [by previous demonstration]. Therefore AU:AT is as UG is to the line [X] that AU cuts from GD [i.e., AU:AT = UG:X], and EA:AU = EG:GU [by Euclid, VI.3], since angle UAG = angle GAE [by construction]. 
◉ Cum ergo eadem sit proportio EA ad AT sicut EG ad GD, proportio EA ad AT sit compacta ex proportione EA ad AU et AU ad AT. Erit proportio EG ad GD compacta ex eisdem, quare erit compacta ex proportione EG ad GU et GU ad lineam quam secat AU ex GD. Sed est compacta ex proportionibus EG ad GU et GU ad GD. Igitur linea quam secat AU ex GD est linea GD. Igitur AU secat GD in puncto D. 
◉ Therefore, since EA:AT = EG:GD [by previous conclusions], EA:AT is compounded from EA:AU and AU:AT [i.e., EA:AT = (EA:AU):(AU:AT)].⁑ EG:GD will be compounded from these same ratios [i.e., EA:AU and AU:AT, so EG:GD = (EA:AU):(AU:AT), but AU:AT = UG:X, by previous conclusions] so it will be compounded from EG:GU and GU:X [i.e., EG:GD = (EG:GU):(GU:X)]. But it is compounded from EG:GU and GU:GD [i.e., EG:GD = (EG:GU):(GU:GD)]. Therefore the line that AU cuts off from GD is line GD [i.e., GD = X]. Therefore AU cuts GD at point D. 
◉ Producatur ergo a puncto A contingens que sit AH. Erit ergo GAH rectus. Sed GAL medietas anguli DGU. Igitur angulus LAH est medietas anguli DGE, cum illi duo valeant duos rectos. Sed cum angulus TAU sit equalis angulo DGU, erit angulus TAD equalis DGE. Igitur angulus LAH est medietas anguli TAD, et angulus EAL medietas anguli EAT. Igitur angulus EAH medietas anguli EAD, quare AH dividit angulum EAD per equalia, quod est propositum. 
◉ Accordingly, from point A extend tangent AH. GAH will therefore be a right angle. But [angle] GAL is half of angle DGU [by previous conclusions]. Hence, angle LAH is half of angle DGE, since these two [i.e., DGU and DGE] sum up to two right angles [so their halves, GAL and LAH, must sum up to a right angle—i.e., GAH, which is right by construction]. But since angle TAU = angle DGU, angle TAD [adjacent to TAU] = [angle] DGE [adjacent to DGU]. Therefore, angle LAH is half of angle TAD [because LAH is half of DGE, by previous conclusions], and angle EAL is half of angle EAT [by previous conclusions]. Thus, angle EAH [which = EAL + LAH] is half of angle EAD, so AH bisects angle EAD, which is what was proposed. 
◉ Si vero AU, cum sit angulus super punctum A equalis angulo GAE, non cadat super lineam ES extra circulum vel intra, sit ergo equidistans [FIGURE 5.2.22a, p. 578]. Igitur angulus UAG equalis est angulo AGE. Sed idem est equalis angulo GAE, quare angulus GAE equalis est angulo AGE. Igitur EG equalis AE. Similiter, angulus TAD erit equalis angulo ATG, quia coalternus. Sed iam dictum est quod angulus TAD est equalis angulo DGT. Igitur angulus ATG est equalis angulo DGT, et similiter duo anguli ADG, DGT sunt equales. Igitur duo anguli ADG, TAD sunt equales. 
◉ On the other hand, if the angle at point A [i.e., UAG] = angle GAE, and if AU does not fall on line ES either outside or inside the circle, then let it be parallel [figure 5.2.22a, p. 236]. Accordingly, [alternate] angle UAG = [alternate] angle AGE. But the same angle [i.e., UAG] = angle GAE [by supposition], so angle GAE = angle AGE. Thus EG = AE. Likewise, angle TAD = angle ATG, since they are alternate. But it has already been established that angle TAD = angle DGT [by previous conclusions]. Therefore, angle ATG = angle DGT, and by the same token the two angles ADG and DGT are equal. Therefore, the two angles ADG and TAD are equal [as are angles ADG and ATG, so triangle AMD is similar to triangle GMT, and both triangles are isosceles because of the equality of the angles at points A and D and at points G, and T]. 
◉ Sequetur ergo ex hiis quod linea quam secat AU ex DG sit equalis linee AT. Et iam dictum est quod EG equalis AE. Igitur proportio EG ad lineam quam secat AU ex DG sicut AE ad AT. Sed iam dictum est quod AE ad AT sicut EG ad GD. Igitur linea quam secat AU ex DG est GD, et cum TAD sit equalis angulo DGT, erit LAH medietas anguli TAD, sicut dictum est supra, et EAL medietas EAT. Erit ergo EAH medietas anguli EAD, quod est propositum. 
◉ From these conclusions it will therefore follow that the line AU cuts from DG is equal to line AT [i.e., GM = MT, and AM = MD, so AM + MT = GM + MD]. And it has already been established that EG = AE. Hence, EG is to the line AU cuts from DG [i.e., X] as AE is to AT [i.e., EG:X = AE:AT]. But it has already been established that AE:AT = EG:GD. Therefore the line that AU cuts off from DG is GD, and since [angle] TAD = angle DGT, [angle] LAH will be half of angle TAD, as was claimed above, and [angle] EAL [will be] half of [angle] EAT [by previous conclusions]. Accordingly, [angle] EAH will be half of angle EAD, which is what was proposed. 
◉ [PROPOSITIO 23] Amplius, dato circulo cuius G [FIGURE 5.2.23, p. 579] centrum, et dato in eo dyametro GB, et dato E puncto extra circulum, est ducere a puncto E ad dyametrum GB lineam secantem circulum ita quod pars eius a circulo usque ad dyametrum sit equalis parti dyametri interiacentis ipsam et centrum. 
◉ [PROPOSITION 23, LEMMA 5] Moreover, given a circle with G as its center [figure 5.2.23, p. 237], given GB as a diameter within it, and given point E outside the circle, a line can be drawn from point E to diameter GB that cuts the circle in such a way that the segment of that line [extending] from the circle to the diameter [i.e., DZ] is equal to the segment of the diameter between that line and the [circle’s] center [i.e., ZG]. 
◉ Verbi gratia, ducatur a puncto E perpendicularis super dyametrum, et sit EC, et ducatur linea EG. Et sumatur linea QT equalis linee EC, et fiat super QT portio circuli ut quilibet angulus cadens in hanc portionem sit equalis angulo EGB, et compleatur circulus. Et a medio puncto QT ducatur perpendicularis ex utraque parte usque ad circulum. Erit quidem dyameter huius circuli. Et a puncto Q ducatur linea ad hunc dyametrum secans eum in puncto F, et producatur usque ad P punctum circuli ita ut FP sit equalis medietati GB, et ducatur linea PT et linea TF. Et ducatur a puncto P linea equidistans dyametro, que sit PU. Concurrat cum TF in puncto U, et a puncto U ducatur equidistans TQ, que sit UO. Et a puncto T ducatur perpendicularis super PQ, que sit TN, et a puncto T ducatur equidistans PQ, que sit TS, et a puncto U perpendicularis super PQ, que sit UH. Deinde ex angulo BGE secetur angulus equalis angulo QPU, qui sit BGD, et ducatur linea EDZ. Dico quod DZ est equalis ZG. 
◉ For instance, from point E draw EC perpendicular to the diameter, and draw line EG. Take a line QT equal to EC, and on QT form a segment of a circle such that any angle within that segment [e.g., QPT] is equal to angle EGB [by Euclid, III.33], and then complete the circle. From the midpoint [L] of QT extend a perpendicular [FL] in both directions to the circle. This will of course be a diameter of this circle. Then, from point Q draw a line to this diameter to intersect it at point F, and extend it to point P on the circle such that FP is half of GB [by proposition 20, lemma 2], and draw line PT and line TF. From point P draw line PU parallel to the diameter. Let it intersect TF at point U, and from point U draw UO parallel to TQ. From point T draw TN orthogonal to PQ, from point T draw TS parallel to PQ, and from point U draw UH orthogonal to PQ. Then, from angle BGE cut off an angle BGD equal to angle QPU, and draw line EDZ. I say that DZ = ZG. 
◉ Et ducatur a puncto D perpendicularis super BG, que sit DI, et ducatur a puncto D contingens, que sit DK. Palam, cum dyameter FL sit perpendicularis super QT et super OU, et PU sit equidistans ei, erit angulus OUP rectus. Et cum OU dividatur a dyametro per equalia et ortogonaliter, erit FO equalis FU, quare angulus FOU equalis angulo FUO. Sed, cum duo anguli POU, OPU valeant rectum, erit angulus FUP equalis angulo FPU, quare FP equalis FU, et ita equalis FO. Et ita PO equalis BG, et equalis GD, et ita proportio EC ad GD sicut TQ ad PO. [2.189] Sed cum angulus KDG sit rectus equalis angulo GID, et angulus IGD communis, erit triangulus IGD similis triangulo KGD, et erit proportio GD ad DI sicut GK ad KD. Sed angulus KGD equalis angulo OPU, et KDG rectus equalis OUP, et ita triangulus KDG similis triangulo OUP, et proportio KG ad KD sicut OP ad OU. Igitur DG ad DI sicut OP ad OU. Ergo proportio EC ad DI sicut QT ad OU. 
◉ Now, from point D draw DI orthogonal to BG, and from point D draw tangent DK. It is clear that, since diameter FL is perpendicular to QT as well as to OU, and since PU is parallel to that diameter, angle OUP will be a right angle. And since OU is bisected by the diameter [FL] along the orthogonal, FO = FU, so angle FOU = angle FUO.⁑ However, since the [remaining] two angles POU and OPU [of triangle POU] sum up to a right angle, angle FUP = angle FPU [because triangle OFU = triangle PFU, since OU and OP are both bisected by FL, and OP is bisected by FU], so FP = FU, and so it equals FO. And therefore PO = BG [because FP was constructed to be half of BG, and it forms half of PO], and it is also equal to GD [since GD and BG are both radii], so EC:GD = TQ:PO [since TQ = EC, by construction]. 
◉ Sed proportio QT ad OU sicut TF ad FU, cum triangulus TFQ sit similis triangulo OFU. Verum angulus UTS equalis angulo HFU, quia coalternus ei, et angulus UST rectus equalis angulo FHU. Erit triangulus UST similis triangulo HUF, et ita proportio TU ad UF sicut SU ad UH, quare proportio TF ad UF sicut SH ad UH. Sed TN equalis SH, cum sit equidistans ei, et sint inter duas equidistantes. Igitur proportio TF ad UF sicut TN ad UH, quare proportio QT ad OU sicut TN ad UH, et EC ad DI sicut TN ad UH. 
◉ But QT:OU = TF:FU, since triangle TFQ is similar to triangle OFU [and QT and OU and TF and FU are corresponding sides]. However, angle UTS = angle HFU, since it is alternate to it [because PQ and ST are parallel, by construction], and right angle UST = [right] angle FHU [both being right by construction. Consequently] triangle UST will be similar to triangle HUF, so TU:UF = SU:UH, and so [by Euclid, V.18] TF:UF = SH:UH. But TN = SH, since it is parallel to it and since both lie between two parallels [PQ and ST]. Therefore, TF:UF = TN:UH, so QT:OU [which = TF:UF, by previous conclusions] = TN:UH, and EC:DI [which = QT:OU, by previous conclusions] = TN:UH. 
◉ Sed cum angulus GID sit rectus equalis angulo PHU, et angulus IGD equalis angulo HPU, erit triangulus IGD similis triangulo HPU, et proportio ID ad GD sicut HU ad UP, quare proportio EC ad GD sicut TN ad UP. Sed cum angulus CGE sit equalis angulo NPT, et angulus GCE rectus equalis PNT, erit proportio GE ad EC sicut PT ad NT. Igitur proportio GE ad GD sicut PT ad UP. 
◉ But since right angle GID = [right] angle PHU, and since angle IGD = angle HPU [by construction], triangle IGD will be similar to triangle HPU [by Euclid, VI.4], and ID:GD = HU:UP, so EC:GD = TN:UP.⁑ Yet, since angle CGE = angle NPT [by construction], and since right angle GCE = [right angle] PNT, [then triangles CGE and NPT are similar, so] GE:EC = PT:NT. Hence, GE:GD = PT:UP.⁑ 
◉ Et angulus DGE equalis angulo UPT. Igitur triangulus DGE similis triangulo UPT. Igitur angulus GDE equalis angulo PUT. Restat ergo angulus GDZ equalis angulo PUF, et angulus DGZ equalis angulo UPF, quare tertius tertio, et proportio DZ ad ZG sicut UF ad FP. Sed UF equalis FP. Ergo DZ equalis ZG, quod est propositum. 
◉ But angle DGE = angle UPT [because angle QPT was constructed equal to angle EGB, and angle DGB = angle HPU, so the remainders DGE and UPT are equal]. Therefore, triangle DGE is similar to triangle UPT. Accordingly, angle GDE = angle PUT. It therefore follows that angle GDZ [of triangle GDZ] = angle PUF [of triangle PUF], and angle DGZ = angle UPF, so the third [angle DZG] = the third [angle PFU, so the triangles are similar], and DZ:ZG = UF:FP. But UF = FP [by previous conclusions]. Therefore, DZ = ZG, which is what was proposed. 
◉ [PROPOSITIO 24] Amplius, dato triangulo ortogonio ABG [FIGURE 5.2.24, p. 580] cuius angulus ABG rectus, et dato in BG vel AB puncto D, est ducere lineam a puncto D ad latus AG concurrentem in puncto qui sit Q et ex alia parte concurrentem cum alio latere ut ipsa totalis se habeat ad GQ sicut E est ad Z. 
◉ [PROPOSITION 24, LEMMA 6] Furthermore, given a right triangle ABG [figure 5.2.24, p. 238] with right angle ABG, and given point D on either BG or AB, to draw a line from point D to side AG intersecting it at point Q [on one side] and intersecting the other side [at point T] in the other direction such that the sum [of the lengths from D to the respective sides] is to GQ as E is to Z [i.e., (TD + DQ = TQ):GQ = E:Z]. 
◉ Verbi gratia, ducatur a puncto D equidistans AB, que sit DM, et fiat circulus transiens per tria puncta D, M, G. Erit MG dyameter. Et ducatur linea AD, et sit H linea ad quam se habeat AD sicut E ad Z. Et cum angulus DMG sit equalis angulo BAG, secetur ex eo equalis angulo DAG, et sit CMD. Et ducatur MC usque contingat circulum in puncto C, a quo ducatur linea ad dyametrum MG et usque ad circulum ita quod LN sit equalis linee H. Et ducatur linea NG, et linea DN concurrens cum AG in puncto Q. 
◉ For instance, from point D draw DM parallel to AB, and produce a circle passing through the three points D, M, and G. MG will be a diameter [since angle MDG is right, by construction]. Draw line AD, and [by Euclid, VI.12] let H be a line in proportion to which AD is as E is to Z [i.e., AD:H = E:Z]. Since angle DMG = angle BAG, cut from it [an angle] equal to angle DAG, and let it be CMD. Draw MC until it meets the circle at point C, and from that point draw a line to diameter MG and [extend it from point of intersection L] to the circle so that LN = line H [by proposition 20, lemma 2]. Then draw line NG, and [draw] line DN to intersect AG at point Q. 
◉ Cum igitur angulus DMC sit equalis angulo DNC, quia super eundem arcum, erit angulus QNL equalis angulo DAQ, et angulus NQL equalis angulo DQA, quare triangulus NQL similis triangulo DQA. Ergo proportio AQ ad QN sicut AD ad NL. 
◉ Accordingly, since angle DMC = angle DNC, because they are subtended by the same arc, and since angle QNL = angle DAQ [because QNL = DMC, which = DAG, by construction], and since angle NQL = [vertical] angle DQA, then [by Euclid, VI.4] triangle NQL is similar to triangle DQA. Therefore, AQ:QN = AD:NL. 
◉ Verum, cum angulus DMG sit equalis angulo DNG, erit QNG equalis angulo TAQ. Sit T punctus in quo DN concurrit cum AB, et angulus TQA equalis angulo NQG. Erit triangulus TQA similis triangulo NQG, et erit proportio AQ ad QN sicut TQ ad QG. Ergo proportio TQ ad QG sicut AD ad LN. Sed NL equalis H, et AD ad H sicut E ad Z. Igitur TQ ad QG sicut E ad Z, quod est propositum. 
◉ But, since angle DMG = angle DNG [because they are subtended by the same arc DCG, then angle] QNG = angle TAQ [because TAQ = DMG, which = DNG]. Let T be the point where DN intersects AB, and [since] angle TQA = [vertical] angle NQG, [then, by Euclid, VI.4] triangle TQA will be similar to triangle NQG, and AQ:QN = TQ:QG. Therefore, TQ:QG = AD:LN [since AQ:QN = AD:LN, by previous conclusions]. But NL = H, and AD:H = E:Z [by construction]. Therefore, TQ:QG = E:Z, which is what was proposed. 
◉ Potest autem contingere quod a puncto C erit ducere duas lineas similes CN, et tunc erit ducere duas lineas a puncto D similes TQ ut utriusque ad partem quam secat ex AG sit proportio sicut E ad Z, et erit eadem probatio. 
◉ Furthermore [as demonstrated in proposition 20, lemma 2], it is possible for two lines like CN to be drawn, and in that case two lines can be drawn from D equal to TQ such that each of them is to the segment it cuts off from AG as E is to Z, and the proof will be identical.⁑ 
◉ [PROPOSITIO 25] Predictis habitis, dato speculo sperico, erit invenire punctum reflexionis in eo. 
◉ [PROPOSITION 25] With these things established, and given a [convex] spherical mirror, it will be [shown how] to find a point of reflection on it. 
◉ Verbi gratia, sit A [FIGURE 5.2.25, p. 581] centrum visus, B punctus visus, G centrum spere, et ducantur linee AG, BG. Et sumatur superficies in qua sunt hee due linee, et sumatur circulus communis huic superficiei et speculo. Invenietur ergo punctus reflexionis in hoc circulo. 
◉ For instance, let A [figure 5.2.25, p. 239] be a center of sight, B a visible point, and G the center of the sphere [forming the mirror], and draw lines AG and BG. Take the plane within which these two lines lie, and take the [great] circle [that forms the] common [section] of this plane and the mirror. Accordingly, the point of reflection will be found on this circle. 
◉ Et sumatur linea alia MK, et dividatur in puncto F ut FM se habeat ad FK sicut BG ad GA. Et dividatur MK per equalia in puncto O, et ducatur a puncto O perpendicularis, que sit CO, et ducatur a puncto K linea ad CO tenens cum CO angulum equalem medietati anguli BGA, que sit KC. Et a puncto F ducatur linea ad CK, que sit FP, et concurrat cum CO in puncto S ita ut proportio SP ad PK sicut BG ad semidyametrum GD. Et ex angulo BGA secetur angulus equalis angulo SPK, scilicet DGB, et ducantur linee SK, BD. 
◉ Take some other line MK, and divide it at point F such that FM:FK = BG:GA [by Euclid, VI.10]. Bisect MK at point O, and from point O draw perpendicular CO, and from point K draw line KC to CO to form an angle [OCK] with CO that is equal to half of angle BGA. Then, from point F draw line FP to CK, and let it intersect CO at point S so that SP:PK = BG:radius GD [by proposition 24, lemma 6]. From angle BGA cut off an angle equal to angle SPK, i.e., [angle] DGB, and draw lines SK and BD. 
◉ Erit igitur proportio BG ad GD sicut SP ad PK, et ita triangulus SPK similis triangulo BGD, et erit angulus SKP equalis angulo BDG. Sed forsan, secundum predictam, poterimus a puncto F ducere aliam lineam ad CK similem SP ut sit proportio eius ad partem quam secabit ex CK sicut SP ad PK, et tunc a puncto K ad OS ducetur alia linea quam SK alium cum CK angulum tenens maiorem vel minorem angulo CKS. Si maior ex hiis angulis non fuerit maior recto, non erit invenire punctum reflexionis. Sit igitur angulus CKS maior recto, et invenitur punctum sic. 
◉ Accordingly, BG:GD = SP:PK [by construction], and so triangle SPK will be similar to triangle BGD [by Euclid, VI.6, because angle SPK = angle DGB, by construction], and [therefore] angle SKP = angle BDG. But, according to what we established earlier [in proposition 24, lemma 6], we may draw from point F to CK another line like SP [i.e., S’P’] such that it is to the segment it will cut off from CK as SP is to PK [i.e., S’P’:P’K = SP:PK], and on that basis a line other than SK [i.e., S’K] will be drawn from point K to OS to form another angle [CKS’] with CK [i.e., other than the original CKS] that is either greater than, or less than angle CKS. If the larger of these angles is not greater than a right angle, no point of reflection will be found [as will be demonstrated below]. Accordingly, let angle CKS be greater than a right angle, and the point [of reflection] is found as follows. 
◉ Erit angulus BDG maior recto. Ducatur contingens NDY, et cum angulus PKO sit minor recto, secetur ex angulo BDG equalis ei, qui sit QDG. Cum igitur angulus SPK sit equalis angulo QGD, erit triangulus FPK similis triangulo QGD, et erit angulus DQB equalis angulo KFS, et triangulus DQB similis triangulo KFS. 
◉ Angle BDG will be greater than a right angle [since BDG = CKS, by previous conclusions, and CKS is greater than a right angle, by stipulation]. Draw tangent NDY, and since angle PKO is smaller than a right angle, cut off angle QDG equal to it from angle BDG. Thus, since angle SPK = angle QGD [i.e., DGB, by construction], triangle FPK will be similar to triangle QGD [by Euclid, VI.4, so angle DQG = angle KFP, and so] angle DQB [adjacent to DQG] = angle KFS [adjacent to KFP], and triangle DQB will be similar to triangle KFS [since angle SKP = angle BDG, and angle QDG = angle PKF, leaving remaining angles QBD and FKS equal]. 
◉ Producatur autem DQ, et a puncto B ducatur perpendicularis super ipsam, que sit BZ. Erit igitur angulus BQZ equalis angulo SFO. Et angulus BZQ rectus equalis angulo SOF, et ita triangulus BQZ similis triangulo SFO. 
◉ Now, extend DQ [beyond Q], and from point B draw perpendicular BZ to it. Accordingly, angle BQZ = angle SFO [since triangles BQD and SKF are similar and BZ and SO are dropped orthogonally to corresponding sides from the corresponding vertices]. Also, right angle BZQ = [right] angle SOF, and so triangle BQZ is similar to triangle SFO [by Euclid, VI.4]. 
◉ Ducatur DZ usque ad punctum I, et sit ZI equalis ZD. Palam ergo quod ZQ ad QB et QB ad QD sicut OF ad FS et FS ad FK, et ex hoc erit ZD ad QD sicut OK ad FK, et ita ID ad QD sicut MK ad FK, et ita IQ ad QD sicut MF ad FK, et IQ ad QD sicut BG ad GA. 
◉ Extend DZ to point I, and let ZI = ZD. It is therefore evident that ZQ:QB and QB:QD are as OF:FS and FS:FK, and from this ZD:QD = OK:FK [by Euclid, V.22], so [by Euclid, V.18] ID:QD = MK:FK [since ID and MK are, by construction, twice ZD and FK, respectively], and so [again, by Euclid, V.17] IQ:QD = MF:FK, and IQ:QD = BG:GA [because MF:FK = BG:GA, by construction]. 
◉ Ducatur autem linea BI et ei equidistans DL. Erit triangulus LDQ similis triangulo BQI, et proportio IQ ad QD sicut IB ad DL. Et cum IZ sit equalis ZD, et BZ perpendicularis, erit BD equalis BI, quare erit BD ad DL sicut BG ad GA. 
◉ Now, draw line BI and [draw line] DL parallel to it. Triangle LDQ will be similar to triangle BQI [by Euclid, VI.4, since angle QBI = alternate angle DLQ, angle BIQ = alternate angle LDQ, and angle BQI = vertical angle DQL], and [so] IQ:QD = IB:DL. And since IZ = ZD [by construction], and since BZ is perpendicular [to ID], BD = BI, so BD:DL = BG:GA [because IQ:QD = BG:GA = IB:DL = BD:DL, by previous conclusions]. 
◉ Ducatur autem a puncto D linea, que sit DH, equalem tenens angulum cum linea LD angulo BGA. Et cum HL et DL concurrant, erunt duo anguli LHD, LDH minores duobus rectis, et ita duo anguli AGH, DHG eis equales sunt minores duobus rectis, quare HD concurret cum GA. Dico quod concurret in puncto A. 
◉ Now, from point D draw line DH to form an angle [HDL] with line LD that is equal to angle BGA. Then, since HL and DL intersect, the two angles LHD and LDH will sum up to less than two right angles, and so the two angles AGH [which = LDH, by construction] and DHG that are equal to these sum up to less than two right angles, so HD will intersect GA. I say that it will intersect at point A. 
◉ Palam quod GDN rectus equalis duobus angulis OCK, OKC, et angulus OKC equalis angulo GDQ. Restat angulus QDN equalis angulo OCK, et ita angulus QDN medietas anguli BGA, et ita medietas anguli HDL. Sed angulus QDB est medietas anguli BDL, quoniam proportio BQ ad QL sicut BD ad DL, cum triangulus DLQ sit similis triangulo BQI, et BD equalis BI. Restat igitur ut angulus NDB sit medietas anguli HDB, et ita BDN equalis NDH. Restat autem BDE equalis angulo HDG. Sed angulus HDG equalis angulo EDA contraposito, quare BDE equalis EDA, et ita D est punctus reflexionis. Ita dico si HD concurrat cum AG in puncto A, quod quidem sic patebit. 
◉ It is evident that right [angle] GDN = the [sum of the] two angles OCK and OKC [in triangle OKC where KOC is a right angle, by construction, and angle NDG is also right, by construction], and angle OKC = angle GDQ [by construction]. It follows that angle QDN = angle OCK [because angles QDN and QDG sum up to a right angle, as do OCK and OKC, and angle QDG = angle OKC in similar triangles QDG and FPK], so angle QDN = half of angle BGA as well as half of angle HDL [since angle QDN = angle OCK = half of angle BGA, by construction, and angle HDL = angle BGA, by construction]. But angle QDB is half of angle BDL, because BQ:QL = BD:DL, since triangle DLQ is similar to triangle BQI [by previous conclusions], and BD = BI [by previous conclusions]. It therefore follows that angle NDB is half of angle HDB, and so [angle] BDN = [angle] NDH. It follows, moreover, that [when radius GD is extended beyond the circle to E, angle] BDE = angle HDG [because NDE and NDG are both right angles, and NDH = NDB, so NDE – NDB = BDE = NDG – NDH = HDG]. But angle HDG = vertical angle EDA, so [angle] BDE = [angle] EDA, and so D is the point of reflection. I say this is so if HD intersects AG at point A, which will be demonstrated as follows. 
◉ Ducatur linea HT equidistans BD. Palam quod angulus BDE equalis est angulo HDG. Sed BDE est equalis angulo HTD, quare HT erit equalis HD. Sed proportio BD ad HT sicut BG ad GH, sicut probat Euclides. Igitur proportio BD ad DH sicut BG ad GH. Sed HD producta concurret cum GA, et fiet triangulus similis triangulo HDL, cum habeat angulum LHD communem, et angulus HDL sit equalis angulo HGA. Igitur proportio HD ad DL sicut HG ad lineam quam secat HD ex GA. Et proportio BD ad DL constat ex BD ad DH et DH ad DL. Igitur constat ex BG ad GH et GH ad lineam quam secat HD ex GA. Sed BD ad DL sicut BG ad GA. Igitur proportio BG ad GA constat ex proportionibus BG ad GH et GH ad lineam quam secat HD ex GA. Sed constat ex proportionibus BG ad GH et GH ad GA. Igitur, GA est linea quam secat HD ex GA, et ita concurret cum ea in puncto A, quod est propositum. [2.209] Si vero angulus CKS non fuerit maior recto, dico quod non fiet reflexio ab aliquo puncto speculi ad visum. 
◉ Draw line HT parallel to BD. It is clear that angle BDE = angle HDG [by previous conclusions]. But [angle] BDE = [alternate] angle HTD, so [triangle HDT is isosceles, and so] HT = HD. But [given that HT is parallel to base BD of triangle BDG, triangles BDG and HTG are similar, so], BD:HT = BG:GH, as Euclid demonstrates [in VI.2]. Thus, BD:DH [which = HT] = BG:GH. But, when it is extended, HD will intersect GA, and it will form a triangle [HGA] similar to triangle HDL [by Euclid, VI.4], since it has angle LHD in common [with triangle HDL], and since angle HDL = angle HGA [by construction]. Therefore, HD is to DL as HG is to the line [X] that HD cuts from GA [i.e., HD:DL = HG:X]. But BD:DL is compounded from BD:DH and DH:DL [i.e., BD:DL = (BD:DH):(DH:DL)]. It is therefore compounded from BG:GH and GH:[X, i.e.,] the line HD cuts off from GA [i.e., BD:DL = (BG:GH):(GH:X)].⁑ But BD:DL = BG:GA [by construction]. Therefore, BG:GA is compounded from BG:GH and GH:[X, i.e.,] the line HD cuts from GA [i.e., BG:GA = (BG:GH):(GH:X)]. But it is compounded from BG:GH and GH:GA [so GA = X]. Therefore, GA is the line HD cuts off from GA, and so it will intersect it at point A, which is what was proposed. 
◉ Si enim dicatur quod potest, sit D punctus reflexionis, et producatur linea AD usque ad H punctum in dyametro BG. Et fiat angulus LDH equalis angulo AGB, et producatur contingens NDY, et fiat angulus QDN equalis medietati anguli AGB. 
◉ For if it is claimed that [such reflection] can [occur], let D be the point of reflection, and draw line AD to point H on diameter BG. Make angle LDH equal to angle AGB, draw tangent NDY, and make angle QDN equal to half of angle AGB. 
◉ Palam quod triangulus HDL similis est triangulo HGA, quare proportio DH ad DL sicut HG ad GA. Sed BD ad DH sicut BG ad GH, quod patebit per HT equidistans BD. Igitur BD ad DL sicut BG ad GA. Sed cum angulus BDE sit equalis angulo HDG, erit angulus BDN medietas anguli BDH. Sed NDQ medietas anguli HDL. Igitur BDQ medietas anguli BDL, quare proportio BQ ad QL sicut BD ad DL. 
◉ It is evident [from previous conclusions] that triangle HDL is similar to triangle HGA, so DH:DL = HG:GA. But BD:DH = BG:GH, which will be evident from the fact that HT is parallel to BD [by construction]. So BD:DL = BG:GA. However, since angle BDE = [alternate] angle HDG, angle BDN will be half of angle BDH [by previous conclusions]. But [angle] NDQ is half of angle HDL [by previous conclusions]. Therefore, [angle] BDQ is half of angle BDL [by previous conclusions], so BQ:QL = BD:DL [by Euclid, VI.3]. 
◉ Ducatur a puncto B equidistans DL, et sit BI, et concurrat DQ cum ea in puncto I. Et dividatur DI per equalia in puncto Z, et ducatur BZ. Erit triangulus BQI similis triangulo QDL. Igitur BQ ad QL sicut BI ad DL, et ita BI equalis BD. Et IQ ad QD sicut MF ad FK, et ita ID ad QD sicut MK ad FK, et ita DZ ad QD sicut OK ad FK, et ita ZQ ad QD sicut OF ad FK. 
◉ From point B draw BI parallel to DL, and let DQ intersect it at point I. Bisect DI at point Z, and draw BZ. Triangle BQI will be similar to triangle QDL [by Euclid, VI.4]. Therefore, BQ:QL = BI:DL, and so BI = BD [by previous conclusions]. But IQ:QD = MF:FK [by previous conclusions], so ID:QD = MK:FK [by Euclid, V.18], DZ:QD = OK:FK [by Euclid, V.17, since OK = half MK, and IZ = half ID], and ZQ:QD = OF:FK [by Euclid, V.17]. 
◉ Palam quod BZ est perpendicularis. Ducatur donec concurrat cum DG in puncto X, quod quidem possibile est, cum angulus DZX rectus, ZDX minor recto. Et palam quod proportio BG ad GD sicut SP ad PK. Cum ergo angulus CKS dicatur non esse maior recto, dico quod super punctum K fiet maior recto per lineam concurrentem cum CO in puncto a quo ducetur linea ad CK transiens per punctum F retinens proportionem ad partem CK sicut BG ad GD. 
◉ It is evident that BZ is perpendicular [to IQ]. Extend it until it intersects DG at point X, which is possible, because angle DZX is a right angle, while ZDX is less than a right angle. And it is obvious that BG:GD = SP:PK [by construction]. Therefore, since it is claimed that angle CKS is not greater than a right angle, I say that an angle greater than a right angle will be formed at point K by a line that intersects CO at the point from which a line passing to CK through point F is to the segment of CK [cut off by it] as BG:GD [i.e., SP:PK = BG:GD].⁑ 
◉ Verbi gratia, planum, cum angulus QDN sit equalis angulo KCO, erit angulus QDG equalis angulo CKO. Fiat ergo super punctum K angulus equalis BDQ, et ponatur quod linea hunc angulum tenens concurrat cum CO in puncto S, et ducatur SFP. Planum est, cum angulus BZD rectus equalis angulo SOK, erit triangulus BZD similis SOK, et proportio BZ ad BD sicut OS ad SK. Sed QZ ad QD sicut OF ad FK. Erit ergo angulus ZBQ equalis angulo OSF, et angulus QBD equalis angulo FSK, quare triangulus BGD similis triangulo SPK. Igitur proportio SP ad PK sicut BG ad GD, quod est propositum. 
◉ For example, it is clear that, since angle QDN = angle KCO [by previous conclusions], angle QDG = angle CKO. Accordingly, at point K form an angle equal to BDQ [i.e., SKO], and suppose that the line forming this angle intersects CO at point S, and draw SFP. It is evident that, since right angle BZD = [right] angle SOK, triangle BZD will be similar to [triangle] SOK, and BZ:BD = OS:SK. But QZ:QD = OF:FK [by previous conclusions]. Hence, angle ZBQ = angle OSF, and angle QBD = angle FSK, so triangle BGD is similar to triangle SPK. Therefore, SP:PK = BG:GD, which is what was proposed.⁑ 
◉ Amplius, impossibile est quod duo anguli supra MO constitituti sit uterque maior recto. Si enim uterque talium maior fuerit recto, cum supra idem centrum fiat angulus equalis angulo SKM, fiet supra idem centrum alius angulus diversus ab isto quem efficiet supra KM alia linea similis SK. Et ita a puncto D et ab alio puncto illius circuli fiet reflexio, quod est impossibile, cum iam probatum sit quod unus uni visui sit reflexionis punctus, et iam ostensum est quomodo inveniri possit. 
◉ Furthermore, it is impossible for two angles to be erected on MO such that both of them are larger than a right angle. For if both such angles were larger than a right angle, then, since an angle equal to angle SKM may be formed upon the same center, another angle different from this one will be formed upon the same center which will form on KM another line like SK. And so reflection will occur from point D as well as from some other point on this circle, which is impossible, since it has already been demonstrated [in proposition 16 above] that for any one center of sight there is [only] one point of reflection, and it has now been shown how to find it.⁑ 
◉ Duobus autem visibus, licet duo sint puncta reflexionis, tamen unica erit ymago sensuali sillogismo, et unicus ymaginis locus. Et hoc probabimus, quoniam due linee a centris oculorum ad centrum circuli ducte sunt equales. 
◉ Given two eyes, even though there are two points of reflection, there will nonetheless be a single image according to sensededuction, and there will be a single imagelocation.⁑ We will demonstrate this on the assumption that the two lines extending from the centers of the eyes to the center of the circle are equal. 
◉ Si autem situs puncti visi respectu utriusque visus sit idem ut linee a puncto viso ad centra oculorum sint equales, facilis erit probatio, quoniam dyametri visuales secant ex circulo arcus reflexionis, et tenent angulos equales cum linea a puncto viso ad centrum spere ducta, et arcus hanc lineam et dyametros visuales interiacentes sunt equales. Et si sumantur puncta reflexionis, secundum supradictam probationem, arcus circuli interiacentes hec duo puncta et punctum circuli quod est in perpendiculari a puncto viso ducta erunt equales, quod facile patebit iterata superiori probatione. 
◉ Now, if the location of the visible point is the same with respect to both eyes, so that the lines [extending] from the visible point to the centers of the eyes are equal, the proof will be simple, because the visual axes cut an arc of reflection on the circle, and they form equal angles with the line extending from the visible point to the center of the sphere [i.e., the normal], and the arcs lying between this line and the visual axes are equal. And if the points [of reflection] are selected according to the previous proof [in proposition 25 above], the arcs on the circle lying between these two points and the point on the circle that lies on the normal extended from the visible point [to the circle’s center] will be equal, which will be easily demonstrated by repeating the preceding demonstration. 
◉ Et hoc sive puncta reflexionis sunt in eadem superficie reflexionis, sive in diversis; erunt tamen arcus illi equales, et linee ducte a centris oculorum ad puncta reflexionum equales, et linee a puncto viso ad eadem puncta equales. Et linee a centris oculorum ad puncta reflexionum procedentes necessario se secabunt, et evidens est probatio quod super idem punctum perpendicularis a puncto viso ducte erit sectio, et in hoc puncto utrique visui apparebit ymago et una sola, quod est propositum. 
◉ And this is the case whether the points of reflection lie in the same plane of reflection, or in different ones; those arcs will still be equal, the lines extending from the centers of the eyes to the points of reflection will be equal, and the lines [extending] from the visible point to the same points will be equal. In addition, the lines extending from the centers of the eyes to the points of reflection will necessarily intersect one another, and the proof is obvious that the intersection will be at the same point on the normal dropped from the visible point, and at this point one and the same image will appear to both eyes, which is what was proposed.⁑ 
◉ Est autem ordinatio ymaginum sicut ordinatio punctorum visorum. Si enim in re visa sumatur linea a capitibus cuius ducantur due linee ad centrum spere, fiet triangulus in quo continebuntur ymagines omnium punctorum illius linee. Et si sit in linea illa punctus eiusdem situs, ymago puncti remotioris ab eo erit in dyametro remotiori ab eius dyametro, et propinquioris in propinquiori. Et ita observatur pars in ymaginibus sicut fuerit in punctis visis. 
◉ Furthermore, the arrangement of images is the same as the arrangement of the visible points [producing them]. For if a line [forming a crosssection] is taken on a visible object, and if two lines are drawn from its endpoints to the center of the sphere, it will form a triangle within which the images of all the points on that [crosssectional] line will be included. And if there is a point on that line that is identically disposed [with respect to both eyes], the image of a point farther from it will lie on a normal farther from its normal, and one nearer [will appear] on a nearer [normal]. And so, in images [any] portion maintains its [relative] position as it actually exists among the visible points [within that portion]. 
◉ Sumpta autem linea in qua est punctus eiusdem situs, quodlibet punctum illius linee eiusdem situs erit respectu duorum oculorum, secundum modum predictum, et unicam habebit ymaginem propter equalitatem angulorum illius linee cum lineis visualibus. Si autem sumatur linea que angulum quem continent due linee a centris oculorum ad punctum visum dividat per equalia, situs cuiuslibet puncti illius linee, quantumlibet producte, erit idem utrique visui sicut fuit alterius, et idem probationis modus. 
◉ Moreover, given a line on which there is [such] a uniformly disposed point, any point on that line will be uniformly disposed with respect to both eyes, according to the previously discussed way, and it will have a single image because of the equality of angles formed by that line with the visual axes. In addition, if a line is chosen that bisects the angle formed by the two lines from the centers of the eye to the visible point, the location of any point on that line, no matter how far it is extended, will be the same for both eyes, just as was the case for the other line, and the same method of demonstration applies. 
◉ Preter has duas lineas non est sumere eundem observantem situm, unde cum punctum visum comprehendatur in perpendiculari, cadet ymago eius in diversis punctis illius perpendicularis, sed imperceptibiliter a se remotis. Et ymago cuiuslibet puncti a quotcumque videatur oculis semper observat ydemptitatem partis, unde apparet unitas ymaginis, sicut dictum est in visu directo. Quod forme, licet in diversa cadant loca, propter tamen distantiam eorum insensiblem non diversificant apparentiam nisi diversificent partem. Similiter hic, quando remotio puncti ab uno visu modicum maior quam ab alio, erunt loca ymaginum imperceptibiliter remota, unde apparent simul, et ex eis una compacta, que quidem ymaginum loca aliquando non totaliter distant, sed partialiter. 
◉ Aside from these two lines, no other one can be found that maintains the same location, so, when a visible point is perceived on the normal, its image will fall at different points on that normal, but [the two resulting images will lie] an imperceptible distance from each other. Also, the image of any point, no matter how many eyes may view it, always maintains a uniform relative position, so the image appears unified, as has been claimed in the case of direct vision. For, even though they may fall at different places [on the normal], still, because of their insensible distance [from one another], the images do not appear disparate unless their relative positions are disparate. In a similar vein, when the distance of a point from one eye is slightly greater than it is from the other, the imagelocations will [only] be imperceptibly separated, so they appear fused, and one [image] is melded from them, in which case the imagelocations may be partially rather than wholly distinct.⁑ 
◉ In speculis columpnaribus exterioribus, aliquando linea communis superficiei reflexionis et superficiei speculi est linea recta, aliquando circulus, aliquando sectio columpnaris. 
◉ In convex cylindrical mirrors, the common section of the plane of reflection and the mirror’s surface is sometimes a straight line, sometimes a circle, and sometimes a cylindric section [i.e., an ellipse].⁑ 
◉ Cum fuerit linea communis linea recta, erit locus ymaginis in perpendiculari a puncto viso ducta super superficiem speculi tantum distans a linea communi quantum punctum visum ab eadem. Et eadem probatio que dicta est in speculo plano. 
◉ When the common section is a straight line, the imagelocation will lie on the normal extending from the visible point to the surface of the mirror, and that imagelocation lies just as far [below the mirror] from the common section [i.e., the line of longitude] as the visible point lies [above it]. [This is subject to] the same proof as applies to the plane mirror [in proposition 2 above]. 
◉ Cum autem communis linea fuerit circulus, erit aliquando ymaginis locus intra circulum, aliquando extra, aliquando in ipso circulo. Eius rei eadem penitus assignatio que in speculo exteriori sperico. 
◉ However, when the common section is a circle, the imagelocation will sometimes lie inside the circle, sometimes outside it, and sometimes on the circle itself. This phenomenon has the same explanation as applies in the convex spherical mirror [in propositions 1115 above]. 
◉ Si vero communis linea fuerit sectio columpnaris, dico quod ymaginum loca quedam intra speculum, quedam in superficie speculi, quedam extra speculum, que in singulari explanabuntur. 
◉ But if the common section is a cylindric section, I say that some of the imagelocations lie inside the mirror, some on the mirror’s surface, and some outside the mirror. These claims will be explained individually. 
◉ [PROPOSITIO 26] Sit ABG [FIGURE 5.2.26, p. 582] sectio columpnaris, B sit punctus reflexionis, E punctus visus, D centrum visus. Et ducatur a puncto B perpendicularis super superficiem contingentem speculum in puncto B, que sit TBQ, et ducatur a puncto E perpendicularis super superficiem contingentem speculum in puncto K, que sit EKQ. Et linea contingens speculum in puncto B sit CU; linea contingens speculum in puncto K sit KM. Dico quoniam due perpendiculares TB, EQ concurrent. 
◉ [PROPOSITION 26] Let ABG [figure 5.2.26, p. 241] be a cylindric section, B the point of reflection, E the visible point, and D the center of sight. From point B draw a normal to the plane tangent to the mirror at point B, let it be TBQ, and from point E draw perpendicular EKQ to the plane tangent to the mirror at point K [to form the normal dropped from the objectpoint]. Let the line tangent to the mirror at point B be CU; let the line tangent to the mirror at point K be KM. I say that the two normals TB and EQ will intersect. 
◉ Ducantur linee EB, DB, et ducatur linea KB. Palam quoniam KM cadet in figura EKB, et linea BC in figura eadem. Igitur BC secabit EK. Secet in puncto C. Palam quoniam angulus TBK est maior recto, et angulus EKB similiter maior recto, quare TB, EK concurrent. Sit concursus punctus Q. Similiter, DBK maior recto. Igitur DB, EK concurrent. Sit concursus punctus H. Igitur H est locus ymaginis. Dico etiam quod proportio EQ ad QH sicut EC ad CH, et etiam quod QH est maior HB. 
◉ Draw lines EB and DB, and draw line KB. It is obvious that KM will fall within figure EKB, and line BC [will fall] within the same figure. Therefore, BC will intersect EK. Let it intersect at point C. It is evident that angle TBK > right angle [CBT], and angle EKB is likewise greater than right angle [EKM], so TB and EK will intersect [since angles BKQ and KBQ, adjacent to angles EKB and TBK are acute]. Let Q be the point of intersection. Likewise, [angle] DBK is greater than a right angle. Therefore, DB and EK will intersect. Let H be the point of intersection. Accordingly, H is the imagelocation. I say, as well, that EQ:QH = EC:CH, and also that QH > HB. 
◉ Ducatur HF equidistans EB. Palam quoniam angulus EBC equalis est angulo DBU. Est igitur equalis angulo CBH. Restat EBT equalis angulo HBQ, cum sit TBC rectus, et QBC rectus. Cum igitur CB dividat angulum EBH per equalia, erit proportio EC ad CH sicut EB ad BH. 
◉ Draw HF parallel to EB. It is clear that angle EBC = angle DBU [because angle of incidence EBT = angle of reflection DBT by construction]. It is therefore equal to angle CBH [which is vertical to angle DBU]. It follows that [angle] EBT = angle HBQ, since TBC is a right angle, and QBC is a right angle [and angle EBT = TBC – EBC, whereas angle HBQ = CBQ – CBH, and EBC = CBH, so the remainders EBT and HBQ are equal]. Thus, since CB bisects angle EBH, EC:CH = EB:BH [by Euclid, VI.3]. 
◉ Sed angulus EBT est equalis angulo HFB, quare HF, HB sunt equalia. Sed proportio EB ad HF sicut EQ ad QH. Erit ergo EC ad CH sicut EQ ad QH, quod est propositum. Et ex hoc, cum sit proportio EQ ad QH sicut EB ad BH, et EQ sit maior EB, erit QH maior HB, quod est propositum. 
◉ But angle EBT = [alternate] angle HFB [since HF and EB are parallel, by construction], so HF and HB are equal [insofar as they form equal angles with FB]. But EB:HF = EQ:QH [because, by Euclid, VI.4, triangles EBQ and HQF are similar, since angle QHF = alternate angle QEB, and angle HFQ = alternate angle EBQ]. Accordingly [since EC:CH = EB:BH = EB:HF = EQ:QH, by previous conclusions], EC:CH = EQ:QH, which is what was proposed.⁑ On this basis, moreover, because EQ:QH = EB:BH, and because EQ > EB, QH > HB, which is [also] what was proposed. 
◉ Palam ex hoc quod, si supra sectionem GB ducatur perpendicularis super superficiem contingentem sectionem, concurret cum TB. Similiter, quecumque ducatur supra sectionem AB concurret cum TB. Et hec quidem patent cum punctus visus non fuerit in perpendiculari visuali. Palam enim ex superioribus quod unius solius puncti forma per perpendicularem accedit ad speculum et secundum eandem reflectitur, et est punctus perpendicularis existens in superficie visus, punctus enim ultra visum sumptus non potest reflecti super hanc perpendicularem quia non potest accedere ad speculum super perpendicularem propter predictam rationem. Et similiter non poterit reflecti ab alio puncto speculi quam a puncto perpendicularis, quia accideret duas perpendiculares concurrere et effici triangulus cuius duo anguli recti, sicut supra patuit. 
◉ From this it is evident that, if a perpendicular is dropped to a plane tangent to segment GB [of the mirror], it will intersect TB. By the same token, any [perpendicular] dropped to segment AB [of the mirror] will intersect TB. And these conclusions are evident when the visible point does not lie on the visual axis. For [when it does] it is clear from earlier discussions [in proposition 8 above] that the form of one single point reaches the mirror orthogonally and is reflected back along the same line, and this point [whose form reaches the mirror along the] normal lies on the surface of the eye, for a point taken outside the eye cannot be reflected along this normal because it cannot reach the mirror along this normal according to the aforementioned reasoning [i.e., that the body of the eye gets in the way]. By the same token, it could not be reflected from a point on the mirror other than from a point on the normal, because [in that case] two normals would happen to intersect and form a triangle with two right angles, just as was shown above [in proposition 8]. 
◉ [PROPOSITIO 27] Amplius, sumatur sectio columpnaris [FIGURE 5.2.27, p. 583], et sumatur in ea punctus A, et ducatur contingens sectionem, que sit AT, et sumatur perpendicularis super AT intra speculum, que sit DA. 
◉ [PROPOSITION 27] Furthermore, take [some] cylindric section [figure 5.2.27, p. 242], select point A on it, draw tangent [E]AT to the section [at point A], and within the [section on the] mirror take DA perpendicular to [E]AT. 
◉ Palam quod AD dividit sectionem in duas partes in quarum utraque est punctus unicus cuius puncti contingens erit equidistans AD. Sit ergo G cuius contingens concurrat cum AD in puncto H, et ducatur perpendicularis super hanc contingentem, que sit QG, et hec quidem necessario concurret cum HD, sicut ostensum est in precedenti figura. Sit concursus in puncto D, et ducatur linea GA usque ad P, et ducatur linea QA. Igitur angulus QAH aut est equalis angulo HAP, aut maior, aut minor. 
◉ It is obvious that AD divides the [cylindric] section into two portions, in either of which there is a single point to which the tangent will be parallel to AD. Accordingly, let G [be a point in one of the two portions of the section] to which the tangent intersects AD at point H, and to this tangent draw perpendicular QG, which will necessarily intersect HD, as was shown in the preceding figure [i.e., 5.2.26]. Let D be the point of intersection, draw line GA to P, and draw line QA. Accordingly, angle QAH is equal to, greater than, or smaller than angle HAP. 
◉ Sit equalis. Procedet igitur forma puncti Q ad A et reflectetur ad P, qui sit visus, et locus ymaginis erit punctus sectionis columpnaris, scilicet G. 
◉ Let it be equal. The form of point Q will therefore reach A and will be reflected to P, which is the center of sight, and the imagelocation will be a point on the cylindric section, i.e., G [because G is where line of reflection PA intersects normal QD]. 
◉ Si vero supra punctum Q sumatur aliquod punctum, ut punctum F, erit quidem angulus FAH minor angulo HAP. Fiat ei equalis NAH. Concurret quidem NA cum GQ intra columpnam. Sit in puncto K. Palam ergo quod ymago puncti F erit in puncto K, et ymagines omnium punctorum ultra punctum Q intra columpnam. 

◉ Si vero intra Q et T sumatur punctum aliquod, ut punctum C, erit angulus CAH maior angulo HAP. Fiat ei equalis HAM. Palam quod MA cadet supra GQ, et extra sectionem. Sit in puncto O. Erit igitur ymago C in puncto O, et omnium punctorum T et Q interiacentium ymagines erunt extra sectionem inter T et G. 
◉ But if some point, such as C, is taken between Q and T, angle CAH > angle HAP. Make it equal to HAM. It is clear that [line of reflection] MA will fall on [normal] GQ, but outside the [cylindric] section. Let it [do so] at point O. Thus, the image of point C will lie at point O, and the images of all points lying between T and Q will lie outside the [cylindric] section between T and G. 
◉ Si autem angulus QAH fuerit minor angulo HAP, secetur ex eo equalis, et sit HAN. Palam quod ymago Q erit in puncto K, et omnium punctorum superiorum ymagines erunt infra sectionem. Si vero inferius sumatur C punctum, ut angulus CAH sit equalis angulo HAP, erit ymago C in sectione, et omnes inter C et Q intra, omnes inter C et T extra. 
◉ Moreover, if angle QAH < angle HAP, then cut [an angle] from HAP equal to it, and let it be HAN. It is evident that the image of Q will lie at point K, and the images of all points above [Q] will lie within the [cylindric] section. But if point C is taken below [Q] so that angle CAH = angle HAP, the image of C will lie on the [cylindric] section, [the images of] all [points] between C and Q [will lie] inside [the cylinder], and [the images of] all [points] between C and T [will lie] outside [the cylinder]. 
◉ Si vero angulus QAH fuerit maior angulo HAP, fiat ei equalis HAM [FIGURE 5.2.27a, p. 583]. Palam quod MA secabit sectionem. Et secet in puncto B, et ducatur contingens super punctum B, que concurrat cum DH in puncto L. Erit autem angulus DLB acutus, et angulus HLB obtusus, et LB concurrens cum HG faciet cum ea acutum. Ducatur perpendicularis a puncto B super LB, que sit SB. Secabit quidem HG, et faciet angulum acutum cum ea, cui angulus contrapositus similiter erit acutus. Et HG secat QA. Sit punctus sectionis U, et facit acutum angulum cum ea super punctum U, quare SB et QU concurrunt. Sit concursus in puncto Z. Palam ergo quod forma puncti Z movebitur ad speculum per ZA, et refertur per AM, et locus ymaginis B. Et ymagines punctorum linee ZS ultra Z erunt intra sectionem, et punctorum citra Z extra sectionem, quod fuit propositum. 
◉ If, however, angle QAH > angle HAP, make [angle] HAM [figure 5.2.27a, p. 242] equal to it. It is clear that MA will intersect the [cylindric] section [at some point beyond A on arc AG]. Let it intersect at point B, draw the tangent to point B, and let it intersect DH at point L. Now, angle DLB will be acute, angle HLB will be obtuse, and, when it intersects HG, LB will form an acute [angle] with it. Draw SB perpendicular to LB at point B. It will intersect HG, it will form an acute angle with it, and its vertical angle will likewise be acute. Let HG intersect QA. Let U be the point of intersection, and it forms an acute angle with QA at point U [i.e., angle HUA is acute], so SB and QU intersect. Let the intersection be at point Z. It is therefore evident that the form of point Z will reach the mirror along ZA, and it is reflected along AM [since QAH = HAM, by construction], and B [will be] the imagelocation. Moreover, the images of points above Z on line ZS will lie inside the [cylindric] section [because both the new ZAH and MAH will be more acute], whereas [the images] of points below Z [on line ZS] will lie outside the [cylindric] section [because both the new ZAH and the new MAH will be more obtuse], which was what was proposed.⁑ 
◉ [PROPOSITIO 28] Amplius, ab uno solo puncto speculi columpnaris fit reflexio ad centrum visus, utpote punctus B [FIGURE 5.2.28, p. 584] reflectitur ad A a puncto G. Dico quod non refertur ad ipsum ab alio puncto speculi quam a puncto G. 
◉ [PROPOSITION 28] Moreover, reflection occurs to a center of sight from only one point on a cylindrical mirror, as, for example, [the form of] point B [figure 5.2.28, p. 243] is reflected to [point] A from point G. I say that it does not reflect to the same point from any point on the mirror other than from point G. 
◉ Quoniam si in superficie reflexionis que est ABG sit totus axis speculi, erit linea communis superficiei speculi et superficiei reflexionis linea longitudinis speculi. Et cum in superficie reflexionis sit centrum visus, punctus visus, punctus reflexionis, et punctus axis in quem cadit perpendicularis, una sola superficies sumi potest in qua sit linea illa longitudinis, sive axis, et puncta A, B, quare non potest fieri reflexio ad A nisi ab aliquo puncto linee longitudinis. Sed iam probatum est quod non potest fieri reflexio ad A ab alio puncto linee longitudinis quam a puncto G, quare in hoc situ ab uno solo puncto speculi fit ad A reflexio. 
◉ For if the entire axis [CED] of the mirror lies in the plane of reflection ABG, the common section of the mirror’s surface and the plane of reflection will be a line of longitude [FGN] on the mirror. And since the center of sight [A], the visible point [B], the point of reflection [G], and the point [E] on the axis where the normal [to the point of reflection] falls [all] lie in the [same] plane of reflection, only one plane can be assumed within which that line of longitude, or the axis, and points A and B lie, so reflection can only occur to A from some point on the line of longitude [FN]. But it has already been demonstrated [in proposition 3] that reflection cannot occur to [point] A from any point other than G on the line of longitude, so in this case reflection occurs to A from only one point on the mirror.⁑ 
◉ Si vero superficies ABG sit equidistans basi columpne, erit linea communis circulus equidistans basi. Et iam patuit quod ab alio puncto illius circuli non potest fieri ad A reflexio. Et si ab alio puncto speculi fiat reflexio, perpendicularis ducta a puncto illo cadet ortogonaliter super axem, et secabit lineam AB in puncto aliquo. A puncto illo ducatur linea ad axem in superficie equidistanti basi columpne. Erit quidem ortogonalis super axem, et ita due perpendiculares efficient cum axe triangulum cuius duo anguli sunt recti, quod est impossibile. Palam ergo quod in hoc situ non refertur B ad A nisi a puncto G. 
◉ But if the plane [of reflection] A’B’G is parallel to the base of the cylinder, the common section [of this plane with the cylinder] will be a circle [GH] parallel to the base. And it has already been demonstrated [in proposition 16] that reflection to [point] A’ cannot occur from any other point on that circle. But if reflection were to occur from some other point on the mirror [outside circle GH], the normal dropped from that point would fall orthogonally to the axis [CED], and it would intersect line A’B’ at some point [K]. From that point draw a line [KLI] to the axis in the plane parallel to the base of the cylinder. It will of course be orthogonal to the axis, and so two perpendiculars [KE and KI] will form with the axis a triangle [KEI] two of whose angles [KEI and KIE] are right angles, which is impossible. It is therefore evident that in this case B does not reflect to A except from point G.⁑ 
◉ Si vero superficies ABG secet speculum sectione columpnari, dico quod a solo puncto G fit reflexio. 
◉ If, however, plane ABG cuts the mirror according to a cylindric section, I say that reflection occurs from point G only. 
◉ Ducatur a puncto A [FIGURE 5.2.28a, p. 584] superficies equidistans basi columpne, que sit EZI, et a puncto G similiter superficies equidistans basi speculi in qua ducatur ab axe linea ad punctum G, que sit TG. Erit quidem perpendicularis super superficiem contingentem speculum in puncto G. Et concurrat cum AB in puncto K, et ducatur a puncto G linea longitudinis speculi, que sit GZ, et sit axis TQ. Et a puncto B perpendicularis ducatur ad superficiem EZI, que sit BH, et ducantur linee AZ, HZ. Et ducatur a puncto Z in superficie illa ad axem linea que sit ZQ. Erit quidem perpendicularis super axem, cum axis sit perpendicularis super hanc superficiem, et erit perpendicularis super superficiem contingentem speculum in puncto Z. Et concurrat cum linea AH in puncto L. Dico quod forma puncti H refertur ad A a puncto Z. 
◉ From [the center of sight at] point A [figure 5.2.28a, p. 243] produce a plane parallel to the base of the cylinder, let it be EZI, and likewise from point [of reflection] G produce a plane parallel to the base of the mirror [i.e., GSP] and in that plane draw line TG from the [mirror’s] axis [TQ] to point G. This line will therefore be perpendicular to the plane tangent to the mirror at point G. Let it intersect [line] AB at point K, and from point G draw GZ, the line of longitude on the mirror, and let TQ be the axis. Then, from [object]point B draw BH perpendicular to plane EZI, and draw lines AZ and HZ. In that same plane [i.e., EZI] draw line ZQ from Z to the axis. It will be perpendicular to the axis, since the axis is perpendicular to this plane [EZI within which it lies], and it will be perpendicular to the plane tangent to the mirror at point Z. Let it also intersect line AH at point L. I say that the form of point H is reflected to A from point Z. 
◉ Ducatur a puncto A equidistans linee KG, que sit AM, que quidem concurret cum BG. Sit concursus in puncto M. Palam quoniam GZ equidistans linee BH, cum utraque sit ortogonalis super superfices equidistantes, quare linea BGM est in superficie harum linearum. Igitur tria puncta M, Z, H sunt in hac superficie. Sed iterum AM est equidistans KG, et LZ equidistans KG, quoniam GZ equidistans TQ et inter superficies equidistantes. Igitur LZ equidistans AM, quare sunt in eadem superficie, et in ea est linea AH. Igitur in hac superficie sunt tria puncta M, Z, H. Et iam patuit quod sunt in superficie MBH. Igitur in linea communi sunt hiis duabus superficiebus. Igitur HZM est linea recta. 
◉ From point A draw [line] AM parallel to line KG, and it will intersect BG. Let M be the point of intersection. It is evident that [line] GZ is parallel to line BH, since [by construction] both of them are perpendicular to parallel planes [i.e., EZI and GSP], so line BGM lies in the same plane [BGMH] as these lines [GZ and BH]. Accordingly, the three points M, Z, and H lie in this plane. But AM in turn is parallel to KG, and LZ is parallel to KG, because GZ is parallel to TQ and lies between parallel planes [i.e., EZI and GSP]. Therefore, LZ is parallel to AM, so they lie in the same plane [AHZM], and line AH lies in it too. Hence, the three points M, Z, and H lie in this [same] plane. But it has already been shown that they lie in plane M[G]BH. So these two planes [AHZM and MGBH] intersect along a common line [MZH]. Therefore, HZM is a straight line. 
◉ Palam, cum G sit punctus reflexionis, erit angulus AGK equalis angulo KGB, et ita equalis angulo AMG. Sed est equalis MAG, quia coalternus. Igitur AG, MG sunt equales. Sed quoniam GZ est ortogonalis super quamlibet lineam superficiei AZH, erit quadratum MG equale quadratis MZ, GZ. Et similiter quadratum AG equale quadratis AZ, GZ. Erit igitur AZ equalis MZ, quare angulus AMZ est equalis angulo ZAM. Sed est equalis angulo LZH, et angulus ZAM est equalis angulo LZA, quia coalternus. Igitur angulus AZL est equalis angulo LZH, quare forma puncti H accedens ad punctum Z refertur ad punctum A. 
◉ Since G is the point of reflection, it is evident that angle [of reflection] AGK = angle [of incidence] KGB, and so angle AGK = angle AMG [which is alternate to angle KGB between parallels AM and KG]. But it is equal to MAG, because it is alternate to it [between the same two parallels]. Hence, [within isosceles triangle MAG] AG and MG are equal. But since GZ is orthogonal to any line within plane AZH, [then, by the Pythagorean Theorem] MG2 = MZ2 + GZ2 [in right triangle MZG]. By the same token, AG2 = AZ2 + GZ2 [in right triangle AZG]. Hence, AZ = MZ [because AG = MG, by previous conclusions], so angle AMZ = angle ZAM. But [since AM and LZ are parallel, by previous conclusions] angle AMZ is also equal to [alternate] angle LZH, and angle ZAM = angle LZA, since they are alternate angles. Therefore, angle AZL = angle LZH, so, when it reaches point Z, the form of point H is reflected to point A. 
◉ Si ergo dicatur quod ab alio puncto quam a puncto G potest forma B reflecti ad A, illud aliud punctum aut erit in linea longitudinis, que est GZ, aut in alia. Si est in ea, ducatur ab eo perpendicularis, que necessario secabit lineam AK, et erit equidistans linee AM. Et linea ducta a puncto B ad illud punctum necessario concurret cum AM, et erit punctus ille et punctus M in eadem superficie. 
◉ Hence, if it is claimed that the form of B can be reflected to A from some point [D] other than G, that other point will lie either on line of longitude GZ or on some other [line of longitude]. If it lies on GZ, draw from it a perpendicular [DL’], which will necessarily intersect line AK and will be parallel to line AM [because, by supposition, it must lie in a plane of reflection that includes points A and B, and it must be normal to the mirror]. In addition, the line extending from point B to that point [D] will necessarily intersect line AM [because line AM is in the plane of reflection that includes A and B], and [so] that point and point M will lie in the same plane. 
◉ Et linea illa aut cadet super punctum M aut super aliud. Si super punctum M, erit ducere a puncto B ad punctum M duas lineas rectas. Si autem ad aliud punctum linee AM, ducatur a puncto illo linea ad punctum Z, et probatur quod hec linea cum HZ facit lineam rectam, sicut probatum est de linea ZM. Et ita a puncto H erit ducere duas lineas rectas per punctum Z transeuntes in diversa puncta AM cadentes, quod est impossibile. 
◉ Furthermore, that line [BD] will either fall on point M or [will fall] on some other point. If [it falls] on point M, then two [different] straight lines will have been drawn from point B to point M [i.e., BGM and BDM]. On the other hand, if [it falls] to another point on line AM [i.e., N], draw a line from that point to point Z, and it is demonstrated that this line forms a straight line with HZ, as was demonstrated for line ZM. And so two straight lines will have been drawn from point H to pass through point Z and fall at different points on AM, which is impossible. 
◉ Palam ergo quod a nullo puncto linee GZ nisi a G potest B reflecti ad A. Si dicatur quod a puncto extra hanc lineam sumpto, ducatur super punctum illud linea longitudinis speculi, et a puncto circuli EZI in quem cadit hec linea, probatur H reflecti ad A secundum supradictam probationem. Sed iam probatum est quod H a puncto Z reflectitur ad A, et ita impossibile. Restat ergo ut a solo puncto speculi reflectitur B ad A, quod est propositum. 
◉ It is therefore evident that [the form of] B can be reflected to A from no point other than G on line GZ. If it is claimed that [it can be so reflected] from a point taken outside this line, draw the line of longitude on the mirror that [lies] on this point, and from the point on circle EZI where this line [of longitude] falls, it is demonstrated that [the form of] H is reflected to A according to the previous proof. But it has already been demonstrated that [the form of] H is reflected to A from point Z, so [the foregoing conclusion] is impossible [by proposition 16 above]. It therefore follows that [the form of] B is reflected to A from only one point on the mirror, which is what was proposed. 
◉ [PROPOSITIO 29] Amplius, dato puncto B, quod reflectitur ad A, erit invenire punctum reflexionis, et hoc patebit per revolutionem probationis. 
◉ [PROPOSITION 29] Furthermore, given that [the form of] point B is reflected to A, it will be possible to find the point of reflection, and this will be demonstrated by reversing the [previous] proof. 
◉ Ducatur a puncto A [FIGURE 5.2.28a, p. 584] superficies equidistans basi columpne, que quidem secabit columpnam super circulum, qui sit EZI. Et ducatur a puncto B perpendicularis super hanc superficiem, que est BH, et inveniatur in hac superficie punctus a quo fit reflexio H ad A, qui sit Z. Et a puncto Z ducatur linea longitudinis, que sit ZG, et a puncto Z perpendicularis ZL et huic equidistans a puncto A, que sit AM. Et linea HZ producatur usque concurrat cum ea, et sit concursus in puncto M. Et a puncto M ducatur linea ad B, que necessario secabit lineam ZG, cum sit in eadem superficie cum ea. Quoniam cum BH sit equidistans GZ, erit HZM in superficie illarum, et ita MB in eadem, que, si secaverit ZG in puncto G, erit G punctus reflexionis, quod quidem si revolvas probationem predictam videre poteris. 
◉ From point A [figure 5.2.28a, p. 243] produce a plane parallel to the cylinder’s base, and this plane will cut the cylinder along circle EZI. Draw line BH orthogonal to this plane from point B, and find point Z within this plane from which the reflection of [the form of point] H occurs to A [by proposition 25 above]. From point Z draw line of longitude ZG, and from point Z [draw] perpendicular ZL, to which AM [extended] from point A is parallel. Extend line HZ until it intersects the latter, and let M be the point of intersection. From point M draw a [straight] line to B, and it will necessarily intersect line ZG, since it lies in the same plane with it. Consequently, since BH is parallel to GZ, HZM will lie in the same plane with them, and so MB [will lie] in that same plane, and if it intersects ZG at point G, then point G will be the point of reflection, and you can see this if you reverse the previous demonstration.⁑ 
◉ In speculis exterioribus piramidalibus, si linea communis superficiei reflexionis et superficiei speculi fuerit linea longitudinis speculi, erit locus ymaginis sicut assignatus est in speculis planis, et eadem probatio. 
◉ In convex conical mirrors, if the common section of the plane of reflection and the surface of the mirror is a line of longitude on the mirror, the imagelocation will be just as it was determined in the case of plane mirrors, and the same proof [applies]. 
◉ Quod autem non possit esse linea communis circulus palam per hoc quod perpendicularis ortogonaliter cadit super superficiem contingentem speculum in puncto reflexionis, et circulus necessario erit equidistans basi. Superficies vero equidistans basi non erit ortogonalis super superficiem contingentem speculum. 
◉ That the common section cannot be a circle is evident from the fact that the normal [must] fall orthogonally to the plane tangent to the mirror at the point of reflection, and [the plane of] the circle will necessarily be parallel to the [mirror’s] base. But no [plane] parallel to [the plane of] the base will be orthogonal to a plane tangent to the mirror. 
◉ Si vero communis linea fuerit sectio piramidalis, ymagines quedam erunt in superficie speculi, quedam intra speculum, quedam extra. Et idem assignationis modus qui est in speculo columpnari exteriori, et eadem probatio. Et sicut ostensum est in columpnari exteriori, per perpendicularem visualem non reflectitur forma ad oculum nisi puncti superficiei oculi tantum, et hoc ab uno solo speculi puncto, et locus ymaginis eius continuus locis aliarum ymaginum, sicut patuit superius. 
◉ If, however, the common section is a conic section, some images will lie on the surface of the mirror, some inside the mirror, and some outside it. And [all this] is determined by the same method as [was applied] in the case of the convex cylindrical mirror, and the same proof [applies]. Furthermore, just as was shown in the case of the convex cylindrical [mirror], the form of only one point on the surface of the eye is reflected orthogonally along the visual axis to the eye, and it does so from only one point on the mirror, and its imagelocation is continuous with the other imagelocations [surrounding it], as was shown above. 
◉ Restat in hiis speculis declarare quod ab uno solo puncto eius fiat reflexio, quod sic patebit. 
◉ It remains [for us] to show that, in these sorts of mirror [i.e., conical convex], reflection may occur from only one point on the mirror, which we will prove as follows. 
◉ [PROPOSITIO 30] Sit A [FIGURE 5.2.30, p. 585] visus, B punctus visus, G punctus reflexionis, et ducatur super punctum G superficies equidistans basi, que quidem secabit piramidem super circulum, qui sit PG. Et ducantur linee AG, BG, AB, et a puncto G ducatur ad centrum circuli linea, que sit GT. Et conus piramidis sit E, a quo ducatur axis, qui erit ET. Et ducatur perpendicularis super superficiem contingentem speculum in puncto G, que sit HG, que, cum dividat angulum AGB per equalia, cadet super AB. Punctus casus sit Z. 
◉ [PROPOSITION 30] Let A [figure 5.2.30, p. 244] be the center of sight, B the visible point, G the point of reflection, and on point G produce a plane parallel to the base, a plane that will cut the cone along circle PG. Draw lines AG, BG, and AB, and from point G draw line GT to the center of the circle. Let the vertex [of the cone] be E, and from it draw axis ET. Then [in the plane of ETG] draw perpendicular HG to the plane [containing line CG that is] tangent to the mirror at point G, and, since [normal] HG bisects angle AGB, it will fall upon AB. Let Z be the point where it falls [on AB]. 
◉ Et a cono ducatur linea longitudinis speculi ad punctum G, que sit EG, cui linee ducatur equidistans a puncto A, que necessario secabit superficiem circuli GP. Secet in puncto N, et sit AN. Similiter, a puncto B ducatur equidistans eidem EG, scilicet BM, que secet superficiem PG in puncto M. Et a puncto N ducatur equidistans GT, que sit NF, et ducantur linee NG, MG, NM. 
◉ From the vertex draw line of longitude EG on the mirror to point G, and draw a line from point A parallel to this line, and this parallel will necessarily intersect the plane of circle GP. Let it be AN, and let it intersect at point N. Likewise, from point B draw a parallel to EG, i.e., BM, and let it intersect the plane [of circle] PG at point M. From point N, as well, draw NF parallel to GT, and draw lines NG, MG, and NM. 
◉ Palam quod TG secabit NM. Secet in puncto Q. Palam etiam quod MG secabit NF, cum secet ei equidistantem. Sit punctus sectionis F. Et a puncto A ducatur equidistans HZ, que sit AL. Palam quod BG concurret cum AL. Sit concursus L. Deinceps ducatur linea communis superficiei contingenti speculum in puncto G et superficiei circuli PG, que sit GC. Palam quod erit ortogonalis super GT, et similiter super NF. 
◉ It is evident that TG will intersect NM. Let it intersect at point Q. It is also evident that MG will intersect NF, since it intersects [line TG] parallel to it. Let F be the point of intersection. Then, from point A draw AL parallel to HZ. It is clear that BG will intersect AL [since A, B, G, and Z lie in the same plane, and AL is parallel to HZ, by construction]. Let L be the [point of] intersection. Then draw GC, which is the common section of the plane [CGE] tangent to the mirror at point G and the plane of circle PG. It is clear that it will be orthogonal to GT, as well as to NF [because NF was constructed parallel to GT]. 
◉ Sumatur etiam linea communis superficiei contingenti et superficiei reflexionis, que sit GD, que quidem, cum secet GH, secabit AL. Sit punctus sectionis D, et erit ortogonaliter super AL. 
◉ In addition, take GD, which is the common section of the plane tangent [to the mirror at point G] and the plane of reflection [AGBZ], and GD will intersect AL, since it intersects GH [to which AL was constructed parallel]. Let D be the point of intersection, and GD will be perpendicular to AL [since it is perpendicular to ZG, which is perpendicular to plane CGED tangent to the mirror at point G and parallel to AL]. 
◉ Palam ex predictis quoniam NF est equidistans GT, et AL equidistans GH. Igitur superficies in qua sunt NF, AL est equidistans superficiei GTH. Sed EG equidistans BM, quare sunt in eadem superficie, que superficies secat predictas equidistantes, unam super lineam EG, aliam super lineam FL, quare FL equidistans EG. Sed AN equidistans eidem. Igitur FL equidistans AN. 
◉ It is clear from the foregoing that NF is parallel to GT [by construction], and AL is parallel to GH [by construction]. Therefore, the plane containing NF and AL [i.e., FLAN] is parallel to plane G[E]TH. But EG is parallel to BM, so they lie in the same plane, and that plane intersects the aforementioned parallel [planes, i.e., GETH and FLAN], one along line [of longitude] EG, the other along line FL, so FL is parallel to EG. But AN is parallel to that same line [EG, by construction]. Therefore, FL is parallel to AN. 
◉ Verum superficies contingens speculum in puncto G secat superficies easdem equidistantes, unam in linea EG, aliam in linea CD. Igitur CD est equidistans EG. Igitur est equidistans AN et LF, quare erit proportio AD ad DL sicut NC ad CF. 
◉ However, the plane tangent to the mirror at point G intersects the same parallel planes [i.e., GETH and FLAN], one along line [of longitude] EG, the other along line CD. Therefore, CD is parallel to EG. It is therefore parallel to AN and LF, so AD:DL = NC:CF [by Euclid, VI.1, VI.2, and V.11]. 
◉ Palam etiam quod angulus BGZ equalis est angulo ZGA, et etiam angulo GLA, et etiam angulo GAL, quare GAL, GLA equales. Et GA et GL equales, et GD perpendicularis super AL. Erit AD equalis DL. Erit igitur NC equalis CF, et GC perpendicularis. Erit angulus CFG equalis angulo CNG, Erit igitur angulus NGQ equalis angulo MGQ. Igitur a puncto circuli PG, quod est G, potest punctus M reflecti ad N, non impediente piramide. 
◉ It is evident as well that angle [of incidence] BGZ = angle [of reflection] ZGA [by construction], and it also equals [alternate] angle GLA [between parallels HZ and AL], as well as angle GAL [which is alternate to angle AGZ between the same parallels], so [angles] GAL and GLA are equal. In addition [because triangle GAL is isosceles], GA and GL are equal, and GD is perpendicular to AL [by previous conclusions. So] AD = DL. Therefore, NC = CF [since AD:DL = NC:CF, as previously concluded], and GC is perpendicular [to FN. So] angle CFG = angle CNG. Hence, angle NGQ = angle MGQ.⁑ Accordingly, [the form of] point M can be reflected to [point] N from point G on circle PG, barring interference from the cone [itself].⁑ 
◉ Dico igitur quod punctum B a solo G refertur ad A. Si enim dicatur quod a puncto alio potest reflecti illud, aut erit in linea longitudinis, que est EG, aut non. 
◉ I say, then, that [the form of] point B is reflected to A only from G. For if it is claimed that it can be reflected from some other point, that point will either lie on line of longitude EG, or it will not. 
◉ Sit in ea, et sit X [FIGURE 5.2.30a, p. 586], et ab eo ducatur perpendicularis super superficiem contingentem speculum in puncto illo, que quidem perpendicularis erit equidistans ZG, et ita equidistans AL. Erit igitur AL in superficie reflexionis huius perpendicularis, et erit similiter in superficie reflexionis perpendicularis ZG. Igitur ille due superficies reflexionis secant se super lineam AL. Sed secant se super punctum B, quod est impossibile, quoniam B non est in linea AL, quod patet per hoc quoniam FL equidistans BM. Restat ergo ut a nullo puncto linee EG preter quam a G possit reflecti B ad A. 
◉ Let it lie on it, let it be X [figure 5.2.30a, p. 245], and from that point draw the normal to the plane tangent to the mirror at that point, and this normal will be parallel to ZG and therefore parallel to AL. AL will therefore lie in the plane of reflection containing this normal, and it will likewise lie in the plane of reflection containing normal ZG. Hence, those two planes of reflection intersect along line AL. But they intersect on point B, which is impossible, because B does not lie on line AL, which is evident from the fact that FL is parallel to BM.⁑ It therefore follows that [the form of point] B can be reflected to A from no other point on line EG than G. 
◉ Si autem ab alio puncto, sit illud U, et ducatur linea longitudinis EUO, et sumatur superficies equidistans basi transiens per punctum U. Palam quoniam AN secabit hanc superficiem. Sit punctus sectionis Y. Similiter BM secabit eandem. Sit punctus sectionis K, et ducantur linee KU, YU, YK. Et cum superficies illa secet piramidem super circulum transeuntem per U, ducatur a puncto U linea ad centrum huius circuli, que sit RU. Et ducantur linee EK, EY, que quidem secabunt superficiem circuli PG, et sint puncta sectionum I, S. Et ducantur linee IO, SO. 
◉ But if [it were assumed to do so] from some other point [not on line of longitude EG], then let that point be U, draw line of longitude EUO, and take the plane parallel to base [PG of the mirror] passing through point U. It is clear that AN will intersect this plane. Let Y be the point of intersection. BM will intersect the same plane as well. Let K be the point of intersection, and draw lines KU, YU, and YK. Since that plane intersects the cone along a circle passing through U, draw line RU from point U to the center of this circle. Draw lines EK and EY, which [when extended] will intersect the plane of circle PG, and let I and S be the points of intersection. Now, draw lines IO and SO. 
◉ Sicut ergo probatum est de puncto M quod, non impediente piramide, potest reflecti ad N a puncto G, ita probatur de puncto K quod potest reflecti ad punctum Y a puncto U, et est eadem probatio. Et ita angulus RUY est equalis angulo RUK. 
◉ Thus, just as it has been demonstrated [earlier in this proposition] for point M that, barring interference from the cone, it[s form] can be reflected to N from point G, so it is demonstrable for point K that it[s form] can be reflected to point Y from point U, and the proof is the same. And so angle RUY = angle RUK. 
◉ Palam ergo quoniam BK est equidistans EG, et linea communis superficiei BGEK et superficiei circuli PG est linea MG. Igitur linea EK, cum sit in hac superficie et secet superficiem circuli PG, cadet super lineam communem, que est MG. Erit igitur SMG linea recta. 
◉ It is therefore evident that BK is parallel to EG [by construction], and the common section of plane BGEK and the plane of circle PG is line MG. Therefore, since it lies in that plane [i.e., BGEK] and intersects the plane of circle PG, line EK will fall on common section MG. SMG will therefore be a straight line. 
◉ Eodem modo, cum superficies NYEG secet superficiem circuli PG super lineam NG, linea EY concurret cum linea NG. Igitur ING linea est recta. Palam etiam quoniam superficies IOE secat superficiem circuli PG super lineam IO, et secat superficiem huic equidistantem que transit per U super lineam YU. Ergo YU equidistat IO. Similiter superficies SOEK secat superficies illas equidistantes super duas lineas SO, KU. Igitur SO equidistat KU. 
◉ By the same token, since plane NYEG intersects the plane of circle PG along line NG, line EY will intersect line NG [at point I]. Thus, ING is a straight line. It is also evident that plane IOE intersects the plane of circle PG along line IO, and it intersects the plane parallel to this one that passes through U along line YU. Therefore, YU is parallel to IO. Likewise, plane SOEK intersects those parallel planes along the two lines SO and KU. Therefore, SO is parallel to KU. 
◉ Similiter, si sumatur superficies secans speculum super lineam longitudinis EO in qua sunt R, U, O, M, secabit illas superficies equidistantes super duas lineas MO, RU. Igitur hee due linee sunt equidistantes. Igitur angulus SOM equalis angulo KUR, et angulus MOI equalis angulo RUY. Sed iam patuit quod angulus KUR equalis est RUY. Igitur angulus SOM equalis est angulo MOI, quare punctus S potest reflecti ad I a puncto O, non impediente piramide. 
◉ Similarly, if the plane cutting the mirror along line of longitude EO and containing R, U, O, and M is taken, it will intersect those parallel planes along the two lines MO and RU. Therefore, these two lines are parallel. Hence, angle SOM = angle KUR, and angle MOI = angle RUY. But it has already been shown that angle KUR = [angle] RUY. Accordingly, angle SOM = angle MOI, so point S can be reflected to I from point O, barring interference from the cone. 
◉ Sed iam probatum est quod punctum M reflecti potest ad I a puncto G, et ita punctum S, quod est in linea SMG, potest reflecti ad I a puncto G. Igitur punctus S reflectitur ad I a duobus punctis circuli PG, quod est impossibile. Restat ergo ut primum sit impossibile, scilicet quod punctus B reflectatur ad A ab aliquo alio puncto speculi quam a G, quod est propositum. 
◉ But it has already been shown [earlier in this proposition] that [the form of] point M can be reflected to I from point G, and so [the form of] point S, which lies on [straight] line SMG, can be reflected to I from point G. Therefore, [the form of] point S is reflected to I from two points on circle PG, which is impossible. It follows, then, that the main [supposition of this proposition], i.e., that [the form of] point B can be reflected to A from some point other than G on the mirror, is impossible, which is what was proposed. 
◉ [PROPOSITIO 31] Amplius, dato speculo piramidali, est invenire punctum reflexionis. 
◉ [PROPOSITION 31] Now, given a [convex] conical mirror, the point of reflection can be found [as follows]. 
◉ Verbi gratia, sit G [FIGURE 5.2.31, p. 587] conus piramidis, et super ipsum fiat superficies equidistans basi piramidis, que sit MNG. A sit punctus visus, B centrum visus. A et B aut erunt citra illam superficiem; aut ultra; aut in ipsa superficie; aut unum citra, aliud ultra; aut unum in superficie, aliud citra vel ultra. 
◉ For instance, Let G [figure 5.2.31, p. 246] be the vertex of the cone, and at that point produce plane MNG parallel to the base of the cone. Let A be the visible point and B the center of sight. A and B will [both] lie above that plane [i.e., MNG]; or [they will both lie] below it; or [they will both lie] in the plane itself; or one [will lie] above it [while] the other [lies] below it; or one [will lie] in the plane [while] the other [lies] above or below it. 
◉ Sint ultra superficiem, et a puncto A ducatur superficies secans piramidem equidistans basi, et ducatur a puncto G linea ad punctum B, que producta cadet in superficiem ab A ductam, cum sit inter superficies equidistantes. Punctus in quo cadit hec linea sit H. 
◉ Let them lie below the plane, and from point A produce a plane that cuts the cone parallel to the base, and from point G to point B draw a line, which, when extended, will fall on the plane produced from A [through the cone], since it lies between parallel planes [i.e., MGN and HEA]. Let H be the point where this line falls [on that plane]. 
◉ Probatur autem modo supradicto quoniam A refertur ad H ab aliquo puncto circuli piramidis quem efficit superficies secans ducta a punctis A, H. Et inveniatur in circulo illo punctus reflexionis, et sit E. Et ducatur linea AB, et linea longitudinis piramidis GE, axis piramidis GT. 
◉ Now, it is proven according to the previous method [applied in proposition 30 above] that [the form of] A is reflected to H from some point on the circle formed on the cone by the plane [HAT] produced from points A and H. Let the point of reflection be found on that circle [by proposition 25], and let it be E. Then draw line AB, and draw the cone’s line of longitude GE, as well as axis GT of the cone. 
◉ Et ducatur a puncto E linea ad centrum circuli, que quidem cadet super axem, et sit ET, et erit ortogonalis super superficiem contingentem circulum illum in puncto E. Et ductis lineis AE HE, secabit angulum earum per equalia, et dividet lineam AH. Sit punctus divisionis R. 
◉ From point E draw line ET to the center of the circle, and it will fall to the axis, and it will be orthogonal to the plane tangent to that circle at point E. Then, with lines AE and HE drawn in, it will bisect the angle formed by them, and it will divide line AH [in half]. Let R be the point of division. 
◉ Palam quoniam GE, ET efficiunt superficiem secantem lineam AB. Sit punctus sectionis F, et a puncto F ducatur perpendicularis super lineam GE et sit FC, que quidem erit ortogonalis super superficiem contingentem piramidem super lineam GE. Deinde a puncto A ducatur equidistans linee FC, et sit AL. FC autem concurret cum axe in punto K. Et a puncto A ducatur equidistans linee RT, que sit AS, et ducatur a puncto E linea communis superficiei AEH et superficiei contingenti piramidem in linea GE, que sit EO. Cadet quidem ortogonaliter super AS, cum sit ortogonalis super ER. 
◉ It is evident that GE and ET form a plane that cuts line AB. Let F be the point of intersection, and from point F draw normal FC to line GE, and it will be perpendicular to the plane tangent to the cone along line GE. Then, from point A draw AL parallel to line FC. FC, moreover, will intersect the axis at point K. From point A draw AS parallel to line RT, and from point E draw common section EO of plane AEH and the plane tangent to the cone along GE. It will fall orthogonally to AS, since it is orthogonal to ER [to which AS is parallel, by construction]. 
◉ Et ducatur linea BC, que producta necessario concurrat cum linea AL. Sit concursus in puncto L, et ducatur a puncto C linea communis superficiei contingenti et superficiei ABL, que sit CP. Et ducantur linee LS, PO. 
◉ Draw line BC, which, when extended, necessarily intersects line AL [which is parallel to FC, by construction]. Let the intersection be at point L, and from point C draw common section CP of the plane tangent [to the cone along GE] and plane ABL. Draw lines LS and PO. 
◉ Palam quoniam superficies ALS est equidistans superficiei GEK, et linee CE, PO sunt in superficie contingenti, que superficies secat illas superficies equidistantes super duas lineas CE, PO. Igitur CE equidistans PO. 
◉ It is obvious that plane ALS is parallel to plane GEK [because AS is parallel to RT, by construction, AL is parallel to FC, by construction, and FC and RT lie in the same plane], and lines CE and PO lie in the plane tangent [to the cone along GE], that plane intersecting those parallel planes [ALS and GEK] along the two lines CE and PO. Hence, CE is parallel to PO. 
◉ Ducatur autem linea HE donec concurrat cum AS in puncto S. Palam quod linea ES est in superficie HEG, et in eadem est linea BL, et hec superficies secat predictas superficies equidistantes in duabus lineis EC, LS. Igitur EC est equidistans LS. Erit igitur PO equidistans LS, quare proportio AO ad OS sicut AP ad PL. 
◉ Furthermore, draw line HE until it intersects AS at point S. It is clear that line ES lies in plane HEG, and BL lies in the same [plane], and this plane cuts the aforementioned parallel planes [ALS and GEK] along the two lines EC and LS. Therefore EC is parallel to LS. Therefore PO will be parallel to LS [since it is parallel to CE], so AO:OS = AP:PL [by Euclid, VI.2]. 
◉ Sed palam quod angulus HER est equalis angulo REA. Erit angulus ESA equalis angulo EAS, et EO perpendicularis. Erit AO equalis OS. Erit igitur AP equalis PL. Et CP perpendicularis super AL, cum sit perpendicularis super FCK. Ergo CL equalis CA, et angulus CLA equalis angulo LAC. Erit ergo angulus BCF equalis angulo ACF. Igitur A refertur ad B a puncto C, quod est propositum. 
◉ But it is clear that angle [of reflection] HER = angle [of incidence] REA [by construction]. Angle ESA = angle EAS [because angle of reflection REA = alternate angle EAS, and angle of incidence HER = alternate angle ESA], and EO is perpendicular [to AS, so] AO = OS [by Euclid, I.26]. Accordingly AP = PL. And CP is perpendicular to AL, since it is perpendicular to FCK. Thus [triangle CAL is isosceles, so] CL = CA, and angle CLA = angle LAC. Accordingly, angle BCF = angle ACF [because, given that FCK and APL are parallel, angle BCF = alternate angle CLA, and angle ACF = alternate angle LAC]. Therefore, A is reflected to B from point C, which is what was proposed. 
◉ Si vero centrum visus et punctus visus fuerit in superficie MGN [FIGURE 5.2.31a, p. 588], sit unum in puncto M, aliud in puncto N, et ducantur linee MG, NG, MN, et dividatur MGN per equalia per lineam UG. Palam quoniam N a puncto G refertur ad M. Palam etiam quod linea UG et axis piramidis sunt in superficie secante piramidem super lineam longitudinis. 
◉ But if the center of sight and the visible point [both] lie in plane MGN [figure 5.2.31a, p. 247], let the former be at point M, the latter at point N, draw lines MG, NG, and MN, and bisect [angle] MGN with line UG. It is clear that [the form of] N is reflected from G to M.⁑ It is also clear that line UG and the cone’s axis lie in a plane that cuts the cone along a line of longitude [GE]. 
◉ A puncto U ducatur ortogonalis super hanc lineam longitudinis, que sit UE. Et super punctum E ducatur superficies equidistans basi, que secabit piramidem super circulum. Linea communis superficiei UEG et huic circulo sit ET. Palam quoniam cadet super axem et super centrum circuli. 
◉ From point U draw UE perpendicular to this line of longitude. Through point E produce the plane parallel to the base [of the cone], and this plane will cut the cone along a circle. Let ET be the common section of plane UEG and this circle. It is evident that it will fall on the axis as well as on the center of the circle. 
◉ Deinde a puncto M ducatur equidistans linee GE, que quidem in superficie illius circuli cadat in punctum H. Similiter, a puncto N ducatur equidistans GE, que cadat in punctum A. Et ducatur AH, et ET secet eam in puncto R. 
◉ Then, from point M draw a line [MH] parallel to GE, a line that falls at point H on the plane of that circle. Likewise, from point N draw a line [NA] parallel to GE and falling at point A [on the plane of the circle]. Draw AH, and let ET intersect it at point R. 
◉ Palam quoniam MH equidistans GE est in eadem superficie cum ea, que superficies secat superficiem MGN et superficiem HEA super duas lineas MG, HE. Igitur MG est equidistans HE. Similiter, AN, GE sunt in superficie secante illas equidistantes super NG, AE. Igitur NG equidistans AE. Similiter, superficies UGE secat easdem superficies super duas lineas RE, UG. Igitur UG, MG equidistantes HE, RE, quare angulus MGU equalis angulo HER, et angulus UGN equalis angulo REA, et angulus HER equalis angulo REA. Et ita punctus A potest reflecti ad H a puncto E. 
◉ It is obvious that MH, which is parallel to GE, lies in the same plane with it, and this plane [MHEG] intersects plane MGN and plane HEA along the two lines MG and HE. Therefore [because planes MGN and HEA are parallel, by construction], MG is parallel to HE. By the same token, AN and GE lie in the plane [ANGE] that cuts those parallel planes [MGN and HEA] along NG and AE. Hence, NG is parallel to AE. Likewise, plane UGE intersects the same planes [MGN and HEA] along the two lines RE and UG. Thus, UG and MG are parallel to HE and RE [in reverse order], so angle MGU = angle HER, angle UGN = angle REA, and angle HER = angle REA. And so [the form of] point A can be reflected to H from point E. 
◉ Si ergo a puncto A ducatur equidistans UE et alia equidistans RE, et ducatur ME donec concurrat cum linea equidistante UE, et ducantur linee communes, ut prius, et iteretur probatio predicta, patebit quoniam N potest reflecti ad M a puncto E. Erit igitur E punctus reflexionis, quod est propositum. 
◉ Accordingly, if a line is drawn from point A parallel to UE, if another [is drawn] parallel to RE, if ME is extended until it intersects the line parallel to UE, if the common sections are drawn, as before, and if the preceding proof is repeated, it will be clear that [the form of point] N can be reflected to M from point E.⁑ E will therefore be the point of reflection, which is what was proposed. 
◉ Si vero ambo fuerint citra MGN [FIGURE 5.2.31b, p. 588], fiat piramis huic opposita. Et est ut protrahantur linee longitudinis piramidis iam facte, et a puncto A ducatur superficies secans hanc ultimam piramidem, que sit equidistans basi, que quidem secabit piramidem super circulum, que sit YZ. 
◉ Now, if both [the center of sight and the visible point] lie above MGN [figure 5.2.31b, p. 247], construct the cone opposite the original one. To do so, extend the lines of longitude of the previously constructed cone, and through point A pass a plane that cuts this latter cone parallel to the base, and it will cut the cone along circle YZ. 
◉ B aut erit in hac superficie, aut non. Si fuerit, fiat operatio a puncto B. Si non, ducatur linea GB usque dum concurrat cum hac superficie. Et sit concursus in puncto D. Palam quoniam A refertur ad D ab aliquo puncto circuli YZ interiori. Inveniatur punctus ille (sicut deinceps probabimus et docebimus, non ex anterioribus), et sit Z. Et ducentur linee DZ, AZ, et linea PZ dividat illum angulum per equalia. 
◉ B will either lie in this plane, or it will not. If it does, then carry out the procedure from point B. If not, then extend line GB until it intersects this plane. Let the intersection be at point D. It is evident that A is reflected to D from some point inside circle YZ.⁑ Find that point (as we will later prove and explain [in proposition 38 below], it is not among those on the anterior [surface of the cone]), and let it be Z. Lines DZ and AZ will be drawn, and let line PZ bisect the angle [formed by them]. 
◉ Et producatur linea ZG ad aliam piramidem, que quidem perveniet ad superficiem eius, et erit linea longitudinis. Et sit linea ZGE. Palam quoniam superficies PZE secabit lineam AB. Secet in puncto Q, et ducatur a puncto Q perpendicularis super lineam GE, et cadat in punctum E. Et erit perpendicularis super superficiem contingentem piramidem super lineam GE. Et super punctum E fiat superficies equidistans basi, que sit AEH, et ducatur a puncto D linea equidistans ZE, que sit DH, concurrens cum superficie illa in puncto H. Et eidem linee sit equidistans AA. 
◉ Extend line ZG to the other cone, and it will reach its surface to form a line of longitude [on it]. Let it be line ZGE. It is obvious that plane PZE will intersect line AB. Let it intersect at point Q, and from point Q draw a perpendicular to line GE, and let it fall at point E.⁑ It will also be perpendicular to the plane tangent to the cone along line GE. Through point E produce plane A’EH parallel to the base [of the cone], and from point D draw line DH parallel to ZE and intersecting that plane at point H. Let A’A be parallel to that same line. 
◉ Palam quoniam DH est equidistans ZE, et sunt in eadem superficie, que superficies secat superficies equidistantes super duas lineas DZ, HE. Igitur HE DZ sunt equidistantes. Similiter, AZ, AE equidistantes. Et palam quoniam PZ transit per centrum circuli YZ, similiter RET per centrum alterius circuli super quem superficies AEH secat piramidem. Igitur superficies PZER secat duas superficies equidistantes super duas lineas PZ, RE. Igitur PZ equidistans RE, quare angulus AZP equalis angulo AER. Et ita erit angulus AER equalis angulo REH, quare A refertur ad H a puncto E. 
◉ It is clear that DH is parallel to ZE, and they lie in the same plane [DHEZ], which intersects the parallel planes [A’EH and AZD] along the two lines DZ and HE. Accordingly, HE and DZ are parallel. Likewise, AZ and A’E are parallel. And it is evident that PZ passes through the center of circle YZ, and so does RET [pass] through the center of the other circle along which plane A’EH cuts the cone. Therefore, plane PZER intersects the two parallel planes [AZD and A’EH] along the two lines PZ and RE. PZ, then, is parallel to RE, so angle AZP = angle A’ER. And therefore angle A’ER = angle REH, so [the form of] A’ is reflected to H from point E. 
◉ Igitur, si a puncto A protraxerimus equidistantem QE, et aliam equidistantem RE, et lineas communes, sicut supra, et iteraverimus modum probandi predictum, patebit quoniam punctus A refertur ad B a puncto E, quod est propositum. 
◉ Accordingly, if a line is drawn from point A’ parallel to QE, if another [line is drawn] parallel to RE, and if the common sections [are drawn] as before, and if we repeat the previous method of proof, it will be evident that [the form of] point A is reflected to B from point E, which is what was proposed.⁑ 
◉ Si vero centrum visus fuerit in superficie equidistante basi que est supra conum, scilicet G, et punctum visus ultra hanc superficiem, erit invenire punctum reflexionis hoc modo. 
◉ If, however, the center of sight lies in the plane parallel to the [cone’s] base and passing through its vertex, i.e., [point] G, and if the visible point lies below this plane, the point of reflection will be found in the following way. 
◉ Sit enim centrum visus M [FIGURE 5.2.31c, p. 589], punctus visus A, et sit MGN superficies equidistans basi piramidis. Et a puncto A ducatur superficies equidistans basi piramidis, que secabit piramidem super circulum DEK cuius centrum T. Et a puncto M ducatur perpendicularis super hanc superficiem, que sit MH, et ducatur linea HT. Et a puncto A ducatur ad lineam HT intra circulum linea AEQ ut EQ sit equalis QT, secundum supradicta. Et ducatur linea TEI, et a puncto H ducatur equidistans TE et equalis, que sit HB . Et ducantur linee MB, BE. Palam quoniam superfices GTE secabit lineam AM. Sit punctus sectionis F, et ducatur a puncto F perpendicularis super lineam GE cadens in puncto O, que sit FOC. Et ducantur linee MO, AO. Dico quoniam O est punctus reflexionis. 
◉ Let M [figure 5.2.31c, p. 248] be the center of sight, A the visible point, and MGN the plane parallel to the base of the cone. From point A produce a plane parallel to the base of the cone, and it will cut the cone along circle DEK with centerpoint T. From point M draw MH perpendicular to this plane, and draw line HT. From point A draw line AEQ to HT within the circle such that EQ = QT, according to the earlier account [in proposition 23, lemma 5]. Then, draw line TEI, and from point H draw HB parallel and equal to TE. Draw lines MB and BE. It is evident that plane GTE will intersect line AM. Let F be the point of intersection, and from point F draw FOC perpendicular to line GE and intersecting it at point O. Then, draw lines MO and AO. I say that O is the point of reflection. 
◉ Palam quoniam HB equidistans et equalis TE. Igitur HT equidistans et equalis BE. Sed MH equidistans et equalis GT, cum utraque perpendicularis. Igitur HT equidistans et equalis MG. Igitur MG equidistans et equalis BE, quare MB equidistans et equalis GE. 
◉ It is clear that HB is parallel and equal to TE [by construction]. Therefore, HT is parallel and equal to BE [since they are cut by equal and parallel lines HB and TE so as to form a parallelogram]. But MH is parallel and equal to GT, since both are perpendicular [to parallel planes DEK and MGN]. Accordingly, HT is parallel and equal to MG. Therefore, MG is parallel and equal to BE [since BE is parallel and equal to HT, by previous conclusions], so MB is parallel and equal to GE. 
◉ Palam etiam quod angulus QTE equalis angulo QET, et ita equalis angulo AEI. Sed est equalis angulo IEB. Igitur, IEB equalis angulo IEA, quare A refertur ad B a puncto E. Et cum MB equidistans sit GE, si a puncto A ducatur equidistans FOC et equidistans TE, et iteretur figura supradicta et probatio, palam quoniam A refertur ad M a puncto O, et ita propositum. 
◉ It is also clear that angle QTE = angle QET [because QT = EQ, by construction, so triangle EQT is isosceles], and so it is equal to angle AEI [which is the vertical angle of QET]. But it is [also] equal to angle IEB [which is alternate to QTE]. Thus, [angle] IEB = angle IEA, so [the form of] A is reflected to B from point E. And since MB is parallel to GE, if a line parallel to FOC, as well as to TE, is drawn from point A, and if the earlier figure and proof [based on figure 5.2.31, p. 282] are repeated, it is clear that A is reflected to M from point O, and so [we have done] what was proposed. 
◉ Si vero M sit in superficie, et A citra superficiem, fiet piramis alia huic opposita. Et fiat super A superficies equidistans basi huius piramidis, et invenitur in circulo huius superficiei punctus reflexionis ex punctis interioribus. Et ducatur a puncto illo linea ADG, et producatur. Et invenietur punctus, secundum superiora, et idem probandi modus. 
◉ Now, if M lies in the plane [of MGN], and A lies above that plane, then the cone opposite the original one will be produced. Pass a plane through A that [cuts the opposite cone] parallel to its base, and the point of reflection is found among points lying inside the circle formed by this plane. Draw the line to G from that point, and extend it. And the point of reflection will be found according to subsequent analysis [in proposition 38 below], and the same method of proof will apply.⁑ 
◉ Si autem puncta, scilicet centrum visus et punctus visus, ita disponantur ut unum sit citra superficiem coni, aliud ultra, sit unum L [FIGURE 5.2.31e, p. 589], aliud A, superficies coni MGN. 
◉ But if the [relevant] points, i.e., the center of sight and the visible point, are so disposed that one of them lies above the plane of the [cone’s] vertex and the other below it, let one of them be L [figure 5.2.31e, p. 249] and the other A, and let the plane at the vertex be MGN. 
◉ Et ducatur a puncto A superficies equidistans basi secans piramidem super circulum DE cuius centrum T, et ducatur linea LG. Concurret quidem cum superficie AED. Sit concursus K, et in circulo DE inveniatur punctus, qui sit E, ita ut contingens ducta a puncto illo, que sit SE, dividat per equalia angulum quem continent linee KE, AE. 
◉ Through point A extend a plane parallel to the base of the cone and intersecting it along circle DE with center T, and draw line LG. It will intersect plane AED. Let K be the [point of] intersection, and in circle DE find point E such that tangent SE drawn from that point bisects the angle formed by lines KE and AE [by proposition 22, lemma 4]. 
◉ Et a puncto L ducatur linea equidistans GE, que necessario concurret cum linea KE. Sit concursus B. Palam quod L est in superficie GEK, et LB in eadem superficie equidistans GE. Et ducatur linea TEI. Palam quoniam superficies GTE secat lineam LA. Secet in puncto U, a quo ducatur perpendicularis super superficiem contingentem, que sit UOC. Et ducantur linee AO, LO. 
◉ Then, from point L draw a line [LB] parallel to [line of longitude] GE, a line that will necessarily intersect line KE. Let B be the intersection. It is clear that L lies in plane GEK, and LB, being parallel to GE, lies in that same plane. Draw line TEI. It is evident that plane GTE intersects line LA. Let it intersect at point U, and from there draw normal UOC to the plane tangent [to the cone at point O]. Finally, draw lines AO and LO. 
◉ Palam quoniam AES equalis est angulo SEK, et cum angulus IES est rectus, et SET rectus, erit IEA equalis angulo TEK. Et ita angulus AEI equalis est angulo IEB, quare A refertur ad B a puncto E. Si ergo a puncto A ducantur equidistans UO et equidistans IT, et iteretur probatio, patebit quoniam refertur A a puncto O ad L, et ita propositum. 
◉ It is clear that [angle] AES = angle SEK [by construction], and since angle IES is a right angle, and [angle] SET is a right angle, [angle] IEA [which = IES + AES] = angle TEK [which = SET + SEK]. And therefore angle AEI = angle IEB [from which it follows that angle of incidence AET = angle of reflection TEB], so [the form of] A is reflected to B from point E. If, then, a line parallel to UO and a line parallel to IT are drawn from point A, and if the proof [from the previous cases] is repeated, it will be clear that [the form of] A is reflected from point O to L, and so [we have done] what was proposed.⁑ 
◉ Palam ergo quomodo sit invenire punctum reflexionis, et hec que dicta sunt in unico visu intelligenda sunt. In duplici autem visu, idem accidit, quoniam eadem forma et idem locus forme comprehendetur ab utroque oculo, et sicut dictum est in speculo sperico exteriori, forme a duobus oculis comprehense in hiis speculis propter contiguitatem videntur eadem, et aliquando simul sunt in loco, aliquando commiscentur earum loca in parte, aliquando separantur, sed modicum. [2.299] Forma autem que secundum perpendicularem in hiis speculis descendit secundum eandem regreditur, sicut supra patuit, et forma illa ab uno oculo super perpendicularem percipitur ab alio oculo secundum lineam reflexionis. Sed loca formarum continua, unde eadem apparet utrique visui forma. 
◉ It is therefore clear how the point of reflection can be found, and what has been discussed must be understood to apply to a single center of sight. But in [the case of] both eyes, the same thing happens, because the same form and the same location for the form is perceived by each eye, and, as has been claimed in the case of the convex spherical mirror, the forms perceived by both eyes in these sorts of mirror [i.e., conical convex] appear identical because of their proximity; sometimes they share precisely the same [image]location, sometimes their [image]locations overlap, and sometimes they are separated, but only a little bit. 
◉ [PROPOSITIO 32] In speculis spericis concavis, aliquando perpendicularis a puncto viso ducta secat lineam reflexionis, aliquando est equidistans ei. Quando secat, erit locus forme aliquando in speculo, aliquando ultra speculum, aliquando citra. Et cum fuerit locus forme citra speculum, aliquando erit inter visum et speculum, aliquando in centro visus, aliquando citra centrum visus. Et nos hoc demonstrabimus. 
◉ [PROPOSITION 32] In the case of spherical concave mirrors, the normal dropped from the visible point [to the mirror] sometimes intersects the line of reflection, and sometimes it is parallel to it. When it intersects, the imagelocation will sometimes lie on the [surface of the] mirror, sometimes behind the mirror, and sometimes in front of it. And when the imagelocation lies in front of the mirror, it will sometimes lie between the center of sight and the mirror, sometimes at the center of sight [itself], and sometimes beyond the center of sight. And we will demonstrate this [as follows]. 
◉ Sit A [FIGURE 5.2.32, p. 590] centrum visus, D centrum speculi, et fiat superficies super hec puncta, que secabit speculum super circulum, qui circulus sit HBFG. Erit quidem hec superficies superficies reflexionis, quoniam est ortogonalis super quamlibet superficiem contingentem circulum. Et ducatur linea AD, et a puncto A ducatur linea ad circulum maior AD, que sit AE. Et a puncto D ducatur ad circulum equidistans linee AE, que sit DH, et producatur AD usque in puncta B, I, et ducatur linea DE. 
◉ Let A [figure 5.2.32, p. 250] be the center of sight and D the center of the mirror, and through these points produce a plane that will cut the mirror along [great] circle HBFG. This plane will be a plane of reflection, because it is orthogonal to any plane that is tangent to the [great] circle [on the sphere]. Draw line AD, and from point A to the circle draw line AE, which is longer than AD. From point D to the circle draw [line] DH parallel to line AE, extend AD to points B and I [on the circle], and draw line DE. 
◉ Palam quoniam angulus AED est minor recto, quoniam ED dyameter, et quelibet linea in circulo cum dyametro facit angulum acutum. Et super punctum E fiat angulus equalis angulo AED, qui sit DET. Palam quoniam ET cadet intra circulum, et secabit lineam DH. Sit punctus sectionis T. Palam etiam quod angulus ADE maior angulo DET, quia AE maior AD, et ita ET secabit AB. Secet in puncto Z. 
◉ It is evident that angle AED is less than a right angle, because ED is a radius, and [by Euclid, III.31] any line in a circle forms an acute angle with the radius [or any other segment of the diameter]. At point E form angle DET equal to angle AED. It is clear that ET will fall inside the circle [since DET is acute] and will intersect line DH. Let T be the point of intersection. It is also clear that angle ADE > angle DET, so ET will intersect AB. Let it intersect at point Z. 
◉ Deinde a puncto A ducatur ad arcum EH linea que sit AN, et ducatur linea DN, et supra punctum N fiat angulus equalis angulo DNA per lineam NM, que necessario cadet intra circulum et secabit DH. Secet in puncto M. Palam etiam quod AN concurret cum DH extra circulum. Sit concursus L. 
◉ Then, from point A draw line AN to arc EH, draw line DN, and at point N form with line NM an angle [DNM] equal to angle DNA, and this line [NM] will necessarily fall inside the circle [because it cannot be tangent at N] and will intersect DH. Let it intersect at point M. It is clear, as well, that AN will intersect DH outside the circle [since DAN and ADH sum up to less than two right angles]. Let L be the intersection. 
◉ Ducatur etiam a puncto A linea ad arcum EF, que sit AG, et ducatur DG, et sit angulus AGD equalis angulo DGQ. Palam quod QG secabit DH. Sit punctus sectionis Q. Palam etiam quod AG concurret cum DH ex parte F. Sit concursus O. Quod autem GQ cadat inter D et H palam cum arcus quem secat GO ex circulo sit maior arcu GH. Si enim ducatur linea GH, angulus HGD respiciet maiorem arcum angulo DGA. 
◉ In addition, draw line AG from point A to arc EF, draw DG, and let angle AGD = angle DGQ. It is obvious that QG will intersect DH. Let Q be the point of intersection. It is also obvious that AG will intersect DH on the side of F. Let O be the intersection. Moreover, since the arc [GY] that subtends GO within the circle is greater than arc GH, it is evident that GQ falls between D and H. For if line GH is drawn, angle HGD[P] will be subtended by a greater arc [i.e., HP] than angle [P]DGA[Y, i.e., arc PY]. 
◉ Item, a puncto A ducatur ad arcum FB linea AC secans DH in puncto S ut sit CS maior SD, et ducatur DC. Palam quod angulus DCA est acutus. Fiat ei equalis, qui sit DCK. Palam, cum angulus CDS sit maior angulo DCS, CK concurret cum DH. Sit concursus in puncto K. 
◉ Furthermore, from point A draw line AC to arc FB so as to cut DH at point S in such a way that CS > SD, and draw DC. It is clear that angle DCA is acute. Form angle DCK equal to it. Since angle CDS > angle DCS, it is evident that CK will intersect DH. Let K be the point of intersection. 
◉ Palam secundum supradicta quod punctus T movetur ad E, et refertur ad A. Et perpendicularis a puncto T ducta est TD, que perpendicularis est super superficiem contingentem circulum, et est equidistans linee reflexionis, que est AE, unde non concurret cum ea. 
◉ According to the foregoing construction [making angle DET = angle AED], it is clear that [the form of] point T propagates to E and is reflected to A. TD is the normal dropped from point T, and this line, which is perpendicular to the plane tangent to the circle [at H], is [by construction] parallel to AE, the line of reflection, so it will not intersect it. 
◉ Punctum autem Z movetur ad E, et refertur ad A. Et perpendicularis a puncto Z ducta est AZ, que concurrit cum AE in puncto A, unde locus forme puncti Z erit A. 
◉ [The form of] point Z, on the other hand, propagates to E and is reflected to A. AZ is the normal dropped from point Z, and it intersects AE at point A, so the imagelocation for point Z will be [at center of sight] A. 
◉ Punctum vero M movetur ad N, et refertur ad A. Et perpendicularis ducta a puncto M, que est MD, concurrit cum AN in puncto L, quod est ultra speculum, et locus forme puncti M erit L. 
◉ On the other hand, [the form of] point M propagates to N and is reflected to A. Normal MD dropped from point M intersects [line of reflection] AN at point L, which lies behind the mirror, and the imagelocation for point M will be L. 
◉ Forma vero puncti Q movetur ad G, et refertur ad A, et locus eius erit O, qui est ultra visum. 
◉ The form of point Q, however, propagates to G and is reflected to A, and its imagelocation [where line of reflection AG intersects normal QD] will be O, which lies beyond the center of sight. 
◉ Et forma puncti K movetur ad C, et refertur ad A, et perpendicularis ab eo est KD, et locus ymaginis S. 
◉ Finally, the form of point K propagates to C and is reflected to A, and the normal [dropped] from it is KD, and [so] its imagelocation is S [where KD intersects line of reflection CA]. 
◉ Palam igitur ex predictis quod ymaginum quedam ultra speculum, quedam inter visum et speculum, quedem in ipso visu, quedam citra visum, quod est propositum. 
◉ From the foregoing it is thus clear that some of the images lie behind the mirror, some between the center of sight and the mirror, some at the center of sight itself, and some beyond the center of sight, which is what was proposed. 
◉ Amplius, palam quoniam visus adquirit formas sibi oppositas, unde, cum locus ymaginis fuerit ultra speculum aut inter visum et speculum, comprehenditur veritas illius ymaginis. Cum autem perpendicularis a puncto viso ducta fuerit equidistans linee reflexionis, apparebit quidem ymago in puncto reflexionis. Quoniam, cum punctus ille sit sensualis, sumpto puncto eius intellectuali medio, ymago cuiuscumque partis illius puncti sensualis ultra medium sumpte erit ultra speculum; ymago partis citra medium erit inter visum et speculum, et cum totalis forma ex ulterioribus et citerioribus videatur una continua, necessario forma illius puncti sensualis videbitur in ipso speculo in loco reflexionis. 
◉ It is evident, moreover, that the visual faculty grasps forms that face it, so, when the imagelocation lies behind the mirror or between the eye and the mirror, the image is grasped according to how it is actually located. However, when the normal dropped from the visible point is parallel to the line of reflection, the image will appear at the point of reflection. For, since that [visible] point is [a] sensible [spot] represented by the point imagined [to lie] at its center, the image of any portion of that sensible spot taken beyond the midpoint [toward the mirror] will lie behind the mirror, whereas the image of the portion in front of the midpoint [away from the mirror] will lie between the eye and the mirror, and since the whole form appears as a continuum composed of the parts behind and the parts in front [of the mirror], the form of that sensible spot will necessarily appear in an imagelocation on the mirror itself.⁑ 
◉ Verum in ymaginibus quarum locus fuerit in centro visus, non comprehenditur veritas earum, unde sepius error accidit in huiusmodi speculis. Quod autem hoc pateat, erigatur super superficiem speculi lignum perpendicularem minus medietate semidyametri speculi. Et citra caput huius ligni sit centrum visus, et dirigatur visus ad punctum speculi cuius longitudo a ligno maior quam longitudo centri visus a dyametro per lignum transeunte. Videbitur quidem ymago illius ligni ultra visum, nec erit certa comprehensio eius, immo apparebit quasi arcuata, cum non sit. In hiis ergo speculis non comprehenditur veritas ymaginis nisi cuius locus fuerit ultra speculum aut inter visum et speculum. Cum autem fuerit centrum visus in perpendiculari per lignum transeunte, non plene comprehendit formam illius ligni. 
◉ But in the case of images that are located at the center of sight, they are not grasped according to how they are actually located, so [visual] deception often occurs in these sorts of mirrors [i.e., spherical concave]. To make this clear, stand a wooden rod less than half the radius of the mirror in length upright upon the surface of the mirror [along a normal]. Let the center of sight be situated directly above this rod, and look at a spot on the mirror that lies farther from the rod than the center of sight does [from the mirror] along the normal passing through the rod. The image of this rod will appear behind the object itself, but it will not be perceived properly; on the contrary, it will appear bowed when it is not.⁑ Hence, in these sorts of mirrors the image is seen according to how it is actually located only when the imagelocation lies behind the mirror or between the eye and the mirror. But when the center of sight lies on the normal passing through the rod, it does not perceive the form of that rod clearly. 
◉ Si vero visus fuerit in dyametro spere et in centro eius, cum quelibet linea ab eo ducta ad speculum sit perpendicularis super speculum, non comprehendetur forma alicuius puncti nisi puncti portionis circuli interiacentis latera piramidis visualis que a centro circuli intelligitur protendi. Quoniam forma cuiuslibet alterius puncti cadet in speculum super lineam declinatam, et necessario refertur super declinatam, quare linea reflexionis non transibit per centrum, et ita non continget centrum visus. 
◉ On the other hand, if the center of sight is located on a diameter of the sphere and at its center, then, since every line dropped from it to the mirror is perpendicular to the mirror, the form of no point will be perceived except a point within the portion of the circle lying between the edges of the visual cone that is imagined to extend [to the mirror’s surface] from the center of the circle [where the center of sight is located]. For the form of any other point will fall on the mirror along an oblique line, and it is necessarily reflected along an oblique line, so the line of reflection will not pass through the center, and so it will not reach the center of sight.⁑ 
◉ Si vero fuerit visus in dyametro sed non in centro, non comprehendet formam alicuius puncti semidyametri in quo est. Quoniam angulus quem efficient due linee a puncto sumpto in semidyametro et a centro visus in idem speculi punctum non dividetur per perpendicularem ab illo puncto speculi ductam, cum illa perpendicularis tendat ad centrum speculi. Sed formam alicuius puncti alterius semidyametri percipere poterit. 
◉ However, if the center of sight lies on a diameter but not at the center, it will not perceive the form of any point on the radius within which it lies. For the angle that the two lines from the given point on the radius to the center of sight will form at the same point on the mirror will not be bisected by the normal extended from that point on the mirror, since that normal is directed to the center of the mirror. But it can perceive the form of any point on [a] radius other [than the one opposite it on the diameter upon which it lies].⁑ 
◉ [PROPOSITIO 33] Amplius, viso puncto in huiusmodi speculo, cum non fuerit perpendicularis equidistans linee reflexionis, linea a centro speculi ad punctum visum ducta sic se habebit ad lineam ab eodem centro ad locum ymaginis ductam sicut linea a puncto viso ad punctum quem diximus contingentie se habet ad lineam a puncto contingentie ad locum ymaginis ductam. 
◉ [PROPOSITION 33] Now, when a point is viewed in this sort of mirror, if the normal is not parallel to the line of reflection, the line extended from the center of the mirror to the visible point will have to the line extended from that same centerpoint to the imagelocation the same ratio that the line drawn from the visible point to the point we have called the [end]point of tangency has to the line extended from the [end]point of tangency to the imagelocation. 
◉ Verbi gratia, sit E [FIGURE 5.2.33, p. 590] centrum speculi, B punctus visus, A centrum visus, G punctus reflexionis, linea contingentie ZG. ZG aut concurret cum EB, aut erit equidistans ei. 
◉ For example, let E [figure 5.2.33, p. 251] be the center of the mirror, B the visible point, A the center of sight, G the point of reflection, and ZG the line of tangency. Either ZG will intersect EB, or it will be parallel to it. 
◉ Concurrat in puncto T. Linea EB concurret cum AG, et non in puncto G, cum EB, BG sint due linee. Igitur aut concurrent ultra G, aut inter G et A, aut in A, aut citra A. Sit ultra G, et in puncto H. Dico ergo quoniam proportio EB ad EH sicut BT ad TH. 
◉ Let it intersect at point T. Line EB will intersect AG, but not at point G, since EB and BG are two [distinct] lines. Therefore, they will intersect behind G, or between G and A, or at A, or in front of A. Let [the intersection be] behind G, at point H. Accordingly, I say that EB:EH = BT:TH.⁑ 
◉ Ducatur perpendicularis EG, et a puncto H ducatur equidistans linee BG, que concurret cum EG. Sit concursus L, et a puncto B ducatur equidistans GH, que necessario concurret cum ZT. Sit concursus Q. 
◉ Draw normal EG, and from point H draw [line HL] parallel to line BG, so it will intersect EG. Let L be the intersection, and from point B draw [line BQ] parallel to GH, so it will necessarily intersect ZT. Let Q be the intersection. 
◉ Palam quoniam angulus BGE est equalis angulo AGE. Sed angulo BGE equalis est angulo GLH, et angulus AGE equalis angulo LGH. Igitur LH equalis est GH. Similiter, angulus BGQ equalis est AGZ, et angulus AGZ equalis angulo GQB, et ita BQ equalis BG, quare proportio BG ad HL sicut BQ ad HG. 
◉ It is evident that angle BGE = angle AGE [by construction]. But angle BGE = [alternate] angle GLH, and angle AGE = [vertical] angle LGH [so triangle LGH is isosceles]. Therefore, LH = GH. Likewise, angle BGQ = [angle] AGZ [by construction], and angle AGZ = [alternate] angle GQB [making triangle GBQ isosceles], and so BQ = BG, from which it follows that BG:HL = BQ:HG [because BG = BQ, and HL = HG]. 
◉ Sed quoniam angulus GHT equalis est angulo TBQ, erit triangulus TBQ similis triangulo GHT. Igitur proportio QB ad HG sicut BT ad TH, et ita BG ad HL sicut BT ad TH. Sed cum triangulus BGE sit similis triangulo HEL, erit proportio BG ad HL sicut EB ad EH, et ita EB ad EH sicut BT ad TH, quod est propositum. 
◉ But since angle GHT = [alternate] angle TBQ, triangle TBQ will be similar to triangle GHT [because they have two equal corresponding angles, i.e., angles TBQ and GHT and angle BTQ and vertical angle GTH]. Thus, QB:HG = BT:TH [by Euclid, VI.4], and so BG:HL = BT:TH [because we have established that BG:HL = QB:HG, and QB:HG = BT:TH]. But since triangle BGE is similar to triangle HEL [HL and GB being parallel, by construction], then BG:HL = EB:EH, and so EB:EH = BT:TH, which is what was proposed. 
◉ Eadem erit probatio si locus ymaginis fuerit inter A et G [FIGURE 5.2.33a, p. 591], aut in A [FIGURE 5.2.33b, p. 591], aut ultra A [FIGURE 5.2.33c, p. 591]. 
◉ The same proof will apply if the imagelocation [H] lies between A and G [figure 5.2.33a, p. 251], at A [figure 5.2.33b], or beyond A [figure 5.2.33c, p. 252]. 
◉ Si vero linea contingentie ZG sit equidistans perpendiculari, que est BH [FIGURE 5.2.33d, p. 591], ducatur perpendicularis GE, que, cum sit perpendicularis super ZG, erit perpendicularis super BH. Et erit angulus BEG equalis angulo HEG, et angulus BGE equalis est angulo EGH. Restat triangulus BGE similis triangulo EGH. Igitur proportio BE ad EH sicut BG ad GH, quod est propositum, quia in hoc casu non potest sumi aliud punctum contingentie quam punctus G, eo modo quo punctum contingentie supra appellavimus. 
◉ If, however, line of tangency ZG is parallel to normal BH [figure 5.2.33d, p. 252], draw perpendicular GE, which, since it is perpendicular to ZG, will be perpendicular to BH. And [right] angle BEG = [right] angle HEG, and angle BGE = angle EGH. It follows that triangle BGE is similar to triangle EGH. Therefore, BE:EH = BG:GH, which is what was proposed, because in this case, no [end]point of tangency other than G can be assumed, according to the way we defined the [end]point of tangency earlier [in proposition 6, where the endpoint of tangency lies at the intersection of the normal dropped from the objectpoint and the line tangent to the point of reflection]. 
◉ [PROPOSITIO 34] Amplius, sit circulus DGT [FIGURE 5.2.34, p. 592], et A centrum visus intra speculum, E centrum speculi, B punctus visus. Et ducatur dyameter DAG. 
◉ [PROPOSITION 34] Now, let DGT [figure 5.2.34, p. 253] be the [great] circle [cut from the mirror by the plane of reflection], A the center of sight inside the circle, E the center of the mirror, and B the visible point. Draw diameter DAG. 
◉ Si fuerit B in semidymetro EG, poterit esse reflexio ab aliquo puncto semicirculi GTD et ab aliquo puncto semicirculi ei oppositi. Quoniam quocumque puncto semidyametri EG sumpto, si ab eo ducatur linea ad aliquod punctum semicirculi GTD, et a puncto A ad idem punctum ducatur alia linea, ille due linee efficient angulum quem dividet semidyameter ductus a puncto E in illud punctum. Similiter in semicirculo opposito. 
◉ If B lies on radius EG, there can be reflection from some point on semicircle GTD as well as from some point on the semicircle opposite [and below] it. For if a line is drawn to some point [F] on semicircle GTD from some selected point [B] on radius EG, and if another line is drawn from point A to the same point [F on semicircle GTD], those two lines will form an angle that the radius [EF] extended from point E will bisect at that point. The same holds in the case of the opposite semicircle. 
◉ Si vero B fuerit extra dyametrum DAG [FIGURE 5.2.34a, p. 592], ducatur dyameter transiens per B, qui sit TBQ. Dico quoniam B potest reflecti ad A per arcum interiacentem semidyametros in quibus sunt A et B, et similiter per eius oppositum, id est per arcum TD et per arcum GQ, et non poterit reflecti ab aliquo puncto arcus GT vel arcus QD. 
◉ But if B lies outside diameter DAG [figure 5.2.34a, p. 253], draw the diameter passing through B, and let it be TBQ. I say that [the form of] B can be reflected to A from the arc lying between the radii in which A and B lie, and likewise from the arc opposite it, i.e., from arc TD and from arc GQ, but it cannot be reflected from any point on arc GT or arc QD. 
◉ Verbi gratia, sumatur punctum in arcu GT, quod sit K, et ducantur linee AK, KB donec KB cadat super dyametrum DG in puncto O. Cum O et A sint ex eadem parte centri circuli quod est E, perpendicularis ducta a puncto K ad E non dividet angulum OKA, et ita B non refertur ad A a puncto K. Similiter, sumpto alio puncto quod sit F, patebit quoniam perpendicularis EF non dividet angulum AFB, et ita non refertur B ad A a puncto F. 
◉ For instance, take some [potential] point [of reflection] K on arc GT, and draw lines AK and KB until KB intersects diameter DG at point O. Since O and A lie on the same side of E, which is the center of the circle, the normal dropped from point K to E will not divide angle OKA, and so [the form of] B is not reflected to A from point K. Likewise, if another [potential] point [of reflection] F is chosen [on that same arc], it will be obvious that normal EF will not divide angle AFB, and so [the form of] B is not reflected to A from point F. 
◉ Quod autem a puncto arcus TD vel arcus GQ possit fieri reflexio palam per hoc. Sit M punctum arcus DT, et ducantur linee AM, MB; fiet quadrangulum AMBE. Igitur perpendicularis EM dividet angulum AMB. 
◉ That reflection can, however, take place from a point on arc TD or on arc GQ is clear from the following. Let M be a [potential] point [of reflection] on arc DT, and draw lines AM and MB to form quadrilateral AMBE. Therefore, normal EM will divide angle AMB. 
◉ Pari modo, sit H punctus arcus GQ. Linea AH secabit dyametrum TQ in puncto C, et linea HB eundem in puncto B. Et sunt hec duo puncta ex diversis partibus centri, quare linea EH dividet illum angulum. 
◉ By the same token, let H be a [potential] point [of reflection] on arc GQ. Line AH will intersect diameter TQ at point C, and line HB [will intersect] the same [diameter] at point B. And these two points lie on different sides of the centerpoint [E], so line EH will divide that angle. 
◉ Pari modo, si fuerit B in superficie speculi aut extra speculum, dum A sit intra speculum, idem erit modus probandi qui prius. Similiter, si A fuerit in superficie speculi, B intus aut exterius. 
◉ Likewise, if B lies on the surface of the mirror, or outside the mirror, as long as A lies inside the mirror, the same method of proof will apply as before. And the same holds if A lies on the surface of the mirror, and B lies inside or outside it.⁑ 
◉ Verum, si a puncto A ducatur equidistans TE, que sit AP [FIGURE 5.2.34b, p. 593], loca ymaginum reflexarum a punctis arcus TP erunt extra speculum; loca autem ymaginum arcus PD ultra centrum visus, quod est A; loca autem ymaginum arcus QG sunt inter centrum visus et speculum. Et quod supradictum est de locis ymaginum idem intelligendum, ducta AM equidistans linee TQ. 
◉ But if AP is drawn from A parallel to TE [figure 5.2.34b, p. 254], the locations of images reflected from points [such as M] on arc TP will lie outside [and behind] the mirror, whereas the locations of images [reflected from points such as N] within arc PD will lie [at points, such as S] beyond the center of sight A, and the locations of images [reflected from points such as H] within arc QG lie [at points, such as X] between the center of sight and the mirror. And the same thing that has been just said about imagelocations must be understood [to apply] when AM is drawn parallel to line TQ. 
◉ Si vero A fuerit extra speculum, B intra [FIGURE 5.2.34c, p. 593], patebit quod diximus. Ducantur a puncto A linee contingentes circulum GTD, que sint AH, AZ, et ducantur duo dyametri AEG, TEQ, et B in dyametro TEQ. Refertur B ad A ab aliquo puncto arcus TD, sed palam quod non ab aliquo puncto arcus ZD. Igitur ab aliquo puncto arcus TZ, et similiter ab aliquo puncto arcus oppositi TD, scilicet arcus GQ. Sed ab arcu TG vel DQ non fiet reflexio secundum supradictum modum. 
◉ On the other hand, if A lies outside the mirror and B inside it [figure 5.2.34c, p. 254], what we claimed will be evident [as follows]. From point A draw lines AH and AZ tangent to circle GTD, draw the two diameters AEG and TEQ, and let B lie on diameter TEQ. [The form of] B is reflected to A from some point on arc TD, but it is obvious that [it is] not [reflected] from any point on arc ZD. Therefore, [it is reflected] from some point on arc TZ, and likewise from some point on arc GQ opposite TD. However, according to the previous method [of demonstration], reflection will not occur from arc TG or [arc] DQ. 
◉ Si vero B sit extra hunc dyametrum et supra alium, qui similiter sit TEQ, fiet reflexio ab arcu TD, et a sola parte eius TZ, et ab arcu opposito, qui est GQ. Sed ab arcu TG vel DQ non fiet reflexio. 
◉ If, however, B lies outside this diameter [TEQ] and on another, which is likewise [designated] T’EQ’, reflection will occur from arc T’D, but only within portion T’Z on it, and [it will occur] from GQ’, the arc opposite it. Reflection will not occur, however, from arc T’G or [arc] DQ’. 
◉ [PROPOSITIO 35] Amplius, sumpto dyametro circuli in sperico speculo concavo, quilibet punctus illius dyametri quantumcumque producti potest esse locus ymaginum. 
◉ If, however, some point, such as point F, is taken above point Q, angle FAH < angle HAP. Make it equal to [angle] NAH. [Line of reflection] NA will intersect [normal] GQ inside the cylinder. Let it [do so] at point K. It is evident, then, that the image of point F will lie at point K, and the images of all points beyond Q [will lie] inside the cylinder. [PROPOSITION 35] Furthermore, if a diameter is taken on a [great] circle within a spherical concave mirror, any point on that diameter, no matter how far it is extended, can be an imagelocation. 
◉ Verbi gratia, sit AG [FIGURE 5.2.35, p. 594] dyameter circuli AMG, cuius D centrum. Sumatur in hoc dyametro punctus Z, E centrum visus. Dico quod Z potest esse locus ymaginis. 
◉ For instance, let AG [figure 5.2.35, p. 255] be a diameter of circle AMG, whose center is D. Take point Z on this diameter, [and take] E [as] the center of sight. I say that Z can be an imagelocation. 
◉ Verbi gratia, ducatur linea ETZ, T punctus circuli. Ducatur linea DT. Erit angulus ETD acutus. Fiat ei equalis qui sit DTL. Palam quod L refertur ad E a puncto T, et eius ymago erit Z. [2.337] Similiter, sumpto puncto L, patebit quod est locus ymaginis. Ducatur linea EL usque in B punctum circuli, et ducatur linea DB. Erit angulus EBD acutus. Fiat ei equalis, qui sit DBC. Refertur quidem punctus C ad E a puncto B, et locus ymaginis eius erit L, et ita sumpto quocumque alio puncto, eadem erit probatio. 
◉ For instance, draw line ETZ, T being a point on the circle. Draw line DT. Angle ETD will be acute. Form [angle] DTL equal to it. It is clear [from the equality of angles ETD and DTL] that [the form of] L is reflected to E from point T and that its image will be Z. 
◉ Amplius, punctorum qui comprehenduntur in hiis speculis quorumdam ymagines quatuor loca sortiuntur, quorumdam tria, quorumdam dua, quorumdam unum. Punctus cuius ymago in quatuor ceciderit loca a quatuor punctis determinatis refertur, non ab aliis, vel pluribus. Punctus cuius ymago tria sibi usurpat loca a tribus punctis speculi refertur, non a pluribus; cuius duo a duobus punctis; cuius autem ymago in unicum cadit locum poterit esse quod ab uno tantum puncto fit eius reflexio, et poterit esse quod a quolibet circuli determinati puncto, non ab alio. 
◉ Now, among the points that are perceived in these sorts of mirrors, some of their images are distributed in four locations, some in three, some in two, and some in [only] one. A point whose image lies at four locations is reflected from four distinct points, and from no others nor from more [than four]. A point whose image occupies three locations is reflected from three points on the mirror, and from no more than three; one whose [image occupies] two [locations is reflected] from two points [only]; whereas one whose image lies in a single location may have its reflection occur from one point only, and it may [have it occur] from some point on a specific [great] circle, but from no other [point within the mirror]. 
◉ [PROPOSITIO 36] Verbi gratia sit E [FIGURE 5.2.36, p. 594] centrum visus; H sit punctus visus in eodem dyametro; D sit centrum circuli. Ducatur dyameter ZEHA. Aut ED est equalis DH, aut non. 
◉ [PROPOSITION 36] For example, let E [figure 5.2.36, p. 255] be the center of sight, let H be a visible point on the same diameter, and let D be the center of the circle. Draw diameter ZEHA. ED is either equal to DH, or not. 
◉ Sit equalis, et super EH ducatur a puncto D perpendicularis dyameter GDB, et ducantur linee HG, GE, HB, BE. Palam quoniam triangulus HGD equalis triangulo EGD, et equalis triangulo HBD et triangulo EBD. Palam quod, cum angulus HGE divisus sit per equalia, H a puncto G refertur ad E, et locus ymaginis eius E. Similiter, H a puncto B refertur ad E, et locus ymaginis eius E. 
◉ Let it be equal, and from point D on EH draw diameter GDB perpendicular [to EH], and draw lines HG, GE, HB, and BE. It is evident that triangle HGD = triangle EGD, and it equals triangle HBD, as well as triangle EBD. Since angle HGE is bisected [by diameter GDB], it is clear that [the form of point] H is reflected from point G to E, and its imagelocation is E [where line of reflection GE intersects normal HD dropped from visible point H]. Likewise, [the form of point] H is reflected to E from point B, and its imagelocation is E. 
◉ Si igitur dyametro ZEHA immoto moveatur semicirculus AGZ per speram, aut solus triangulus HGE, describet quidem punctus G motu suo circulum, et a quolibet puncto illius circuli refertur H ad E, et locus ymaginis eius semper erit punctus E, et ita propositum. 
◉ Therefore, if diameter ZEHA remains stationary and semicircle AGZ, or only triangle HGE, is rotated [around AZ, as an axis] throughout [the circumference of] the sphere, point G will describe a circle in the course of its motion, and from any point on that circle [the form of point] H is reflected to [point] E, and its imagelocation will always be point E, and so [we have demonstrated] what was proposed. 
◉ Quod ab alio puncto quam illius circuli non possit fieri reflexio puncti H ad E palam per hoc. Sumatur punctum C. Erit quidem linea EC maior linea EG, et linea HC minor linea HG, quare non erit proportio EC ad HC sicut ED ad DH. Igitur linea DC non dividet angulum ECH per equalia, quare H a puncto C non potest reflecti ad E. Eadem erit improbatio si sumatur C inter G et Z. 
◉ That reflection of [the form of] point H to E cannot take place from any point other than one on that circle [of revolution] is evident from the following. Take point C. Line EC > line EG, and line HC < line HG, so EC:HC ≠ ED:DH [which means that angle ECD ≠ angle DCH, by Euclid, VI.3]. Therefore, line DC will not bisect angle ECH, so [the form of point] H cannot be reflected to E from point C. The same disproof will apply if C is taken between G and Z. 
◉ Si vero ED fuerit maior DH, mutetur figura, et addatur linee EH linea HQ [FIGURE 5.2.36a, p. 595] ut productum ex EQ in QH sit equale quadrato DQ. Erit igitur proportio EQ ad DQ sicut DQ ad HQ, unde EQ ad DQ sicut ED ad DH, sicut probat Euclides. 
◉ But if ED > DH, adjust the figure [accordingly], and add line HQ to line EH [figure 5.2.36a, p. 256] so that EQ,QH = DQ2. Accordingly EQ:DQ = DQ:HQ [by Euclid, VI.17], so EQ:DQ = ED:DH, as Euclid demonstrates [in V.19].⁑ 
◉ Fiat circulus ad quantitatem semidyametri QD, cuius Q centrum, G, B loca sectionis duorum circulorum, et ducantur linee EG, EB, QG, QB, DG, DB, HG, HB. Palam ergo quod erit proportio EQ ad QG sicut QG ad QH, et angulus GQH communis utrique triangulo EQG, HQG. Igitur illi duo trianguli sunt similes. Erit ergo proportio EQ ad QG sicut EG ad GH. Erit igitur proportio ED ad DH sicut EG ad GH, quare linea DG dividet angulum EGH per equalia. 
◉ Produce a circle with a radius of QD having its center at Q, let G and B be the points where the two circles intersect, and draw lines EG, EB, QG, QB, DG, DB, HG, and HB. Accordingly, it is evident that EQ:QG = QG:QH [because radius QD = radius QG, and, as established earlier, EQ:QD = QD:QH], and angle GQH is common to both triangles EQG and HQG [whereas angles QHG and QGE are right]. Therefore, those two triangles are similar. So [by Euclid, VI.4] EQ:QG = EG:GH. Therefore, ED:DH = EG:GH [since we established that EQ:QG = EQ:QD = ED:DH], so line DG will bisect angle EGH [by Euclid, VI.3]. 
◉ Unde punctus H a puncto G refertur ad E, et locus ymaginis eius punctus E. Similiter, H a puncto B refertur ad E, et locus ymaginis est E. 
◉ Consequently, [the form of] point H is reflected to E from point G, and its imagelocation is point E. Likewise, [the form of point] H is reflected from point B to E, and its imagelocation is E. 
◉ Si ergo moveatur triangulus EGH, punctis E, H immotis, punctus G describet in spera circulum a quolibet puncto cuius refertur H ad E, et semper erit locus ymaginis E. 
◉ Hence, if points E and H are held stationary and triangle EHG is rotated [around QE, as axis], point G will describe within the sphere a circle from any point of which [the form of] H is reflected to E, and E will always be the imagelocation. 
◉ Et quod ab alio puncto quam illius circuli non possit H reflecti ad E palam, ut prius. Si enim sumatur C inter G et A, erit EC maior EG, et HC minor HG, nec erit proportio EC ad HC sicut ED ad DH, et ita DC non dividit angulum ECH per equalia. Similiter, si C sumatur inter G et Z poterit improbari. 
◉ Moreover, as before, it is clear that [the form of point] H cannot be reflected to point E from any point [within the sphere] other than one on that circle. For if C is taken between G and A, EC > EG, and HC < HG, so EC:HC ≠ ED:DH, and so [by Euclid, VI.3] DC does not bisect angle ECH. Likewise, if C is taken between G and Z, [then the supposition that reflection can occur from C] can be disproved. 
◉ Et ita propositum, notandum tamen quod E est punctus intellectualis, et circulus ille cuius E est polus est circulus intellectualis, et H punctus intellectualis. Unde quod dictum est secundum geometricam demonstrationem est intelligendum non secundum visus probationem, cum intellectualia visum lateant. Sed quoniam forma H continua videtur formis aliorum punctorum, videbitur quidem a visu forma cuius punctus medius H, et locus puncti medii illius forme erit E, et reflectetur hec forma a loco speculi circularis cuius medium erit circulus predictus, et E polus eius. 
◉ And so what was set out [has been demonstrated], as long as it is borne in mind that E is an imaginary point, and the circle [of revolution] with E as pole is an imaginary circle, and H is an imaginary point. Therefore, what has been claimed according to geometrical demonstration is not to be understood according to ocular proof, because imaginary objects are invisible to sight. But since the form of H appears continuous with the forms of other points [in close proximity within the visible spot represented by H], the form that sight will see is a form whose midpoint is H, and the midpoint of that form [when reflected to E] will be E, and this form will be reflected from a circular segment of the mirror within which the aforementioned circle [created by the rotation of G] will represent the midline with E as its pole. 
◉ Cum autem ED sit maior DH, et in tantum poterit esse maior quod non refertur H ad E a puncto G, sciendum quod nisi fuerit proportio EA ad AH maior quam ED ad DH, non poterit H reflecti ad E. 
◉ Moreover, if ED > DH, insofar as it can be enough longer that [the form of point] H does not reflect to E from point G, it must be understood that, unless EA:AH > ED:DH, [the form of] H cannot be reflected to E. 
◉ Si enim potest reflecti, reflectatur a puncto quod sit G. Erit quidem angulus GDH minor recto, cum respiciat sectionem minorem quarta. Ducatur a puncto G contingens, que necessario concurret cum EA. Sit concursus Q. Erit quidem proportio EQ ad QH sicut ED ad DH (ex [33]), sed maior est proportio EA ad AH quam EQ ad QH. Igitur maior est EA ad AH quam ED ad EH, et ita necessario, si H refertur ad E, erit proportio EA ad AH maior ED ad DH. Palam ergo que dicta sunt cum centrum visus et punctus visus fuerint in eodem dyametro. 
◉ For if it could be reflected, let it be reflected from point G. Angle GDH will be less than a right angle, since it is subtended by an arc less than a quarter of the circle. From point G draw the tangent, which will necessarily intersect EA. Let Q’ be the intersection. By the [33d proposition⁑ we know that] EQ’:Q’H = ED:DH, but EA:AH > EQ’:Q’H. Therefore, EA:AH > ED:DH, and so, by necessity, if [the form of point] H is reflected to E, EA:AH > ED:DH.⁑ The claims that have been made are therefore obvious when the center of sight and the visible point lie on the same diameter. 
◉ [PROPOSITION 37] Amplius, cum punctum visum et centrum visus non fuerint in eodem dyametro, et fuerint extra speculum, non refertur punctus visus ad centrum visus nisi ab uno tantum speculi puncto. 
◉ [PROPOSITION 37] Furthermore, if the visible point and the center of sight do not lie on the same diameter, and if they lie outside the mirror, the [form of the] visible point is reflected to the center of sight from only one point on the mirror. 
◉ Verbi gratia, sit T [FIGURE 5.2.37, p. 595] punctus visus, H centrum visus, D centrum spere, et ducantur linee HD, TD. Superficies quidem HDT secat speram super circulum EBQG. 
◉ For instance, let T [figure 5.2.37, p. 257] be the visible point, H the center of sight, and D the center of the sphere, and draw lines HD and TD. Plane HDT intersects the sphere along circle EBQG. 
◉ Palam quoniam T non refertur ad H nisi ab aliquo puncto huius circuli. Palam etiam quod non refertur ab arcu QG vel BA, secundum modum predictum. Refertur ergo aut ab arcu GB aut AQ. 
◉ It is clear that [the form of] T is not reflected to H except from some point on this circle. According to earlier discussion [in proposition 34 above], it is also clear that it is not reflected from arc QG or BA. It is therefore reflected from either arc GB or AQ. 
◉ Dividatur angulus TDH per equalia per lineam LEDZ, et a puncto E ducatur contingens, que sit KEF. Si puncta T, H fuerint super illam contingentem, non reflectetur T ad H ab aliquo puncto arcus BG. Cum enim a puncto T ducetur linea ad aliquem interiorem punctum huius arcus, linea a puncto H ad idem punctum ducta cadet super ipsum exterius, non interius, et ideo non erit reflexio. 
◉ Bisect angle TDH with line LEDZ, and draw tangent KEF from point E. If points T and H lie on that tangent, [the form of point] T will not be reflected to H from any point on arc BG. For if a line will be drawn from point T to some point on this arc [on the] inner [side of the circle], the line drawn from point H to the same point will fall to it on the outer side [of the circle], not the inner side, and so there will be no reflection.⁑ 
◉ Et quod ab uno puncto arcus AQ tantum fiat reflexio palam erit ex hoc. Ducantur linee TZ, HZ. Cum angulus TDH divisus sit per equalia, TDZ equalis angulo HDZ. 
◉ Moreover, it will be clear from the following that reflection may occur from only one point on arc AQ. Draw lines TZ and HZ. Since angle TDH is bisected, [angle] TDZ = angle HDZ. 
◉ Linee TD HD aut sunt equales, aut non sunt equales. Si sint equales, et DZ communis, erit triangulus TZD equalis triangulo HZD, et angulus TZH divisus per equalia per lineam DZ, et ita T refertur ad H a puncto Z. 
◉ Either lines TD and HD are equal, or they are not equal. If they are equal, and if line DZ is common, then triangle TDZ = triangle HZD, and angle TZH is bisected by line DZ, and so [the form of point] T is reflected to H from point Z. 
◉ Quod ab alio puncto non possit sic constabit. Sumatur punctus O, et ducantur linee TO, HO, et linea ODM dividat angulum illum per equalia. Planum quod TZ minor TO, et HO minor HZ, et proportio TZ ad HZ sicut TL ad LH, et erit proportio TO ad HO sicut TM ad MH. Sed minor est proportio HO ad TO quam HZ ad TZ. Ergo minor HM ad MT quam HL ad LT, quod est impossibile. 
◉ That it cannot [be reflected] from another point [on arc AQ] will be established as follows. Take point O, draw lines TO and HO, and let line ODM bisect the angle [formed by them]. It is clear [by Euclid, III.8] that TZ < TO and that HO < HZ, and [by Euclid, VI.3 it is also clear that] TZ:HZ = TL:LH [since angle HTZ is bisected by ZL, leaving triangles HZL and TZL equal] and TO:HO = TM:MH [since angle TOH is supposedly bisected by OM, leaving triangles HOM and TOM equal]. But HO:TO < HZ:TZ [since TO > HO, and HZ = TZ]. Accordingly, HM:MT < HL:LT, which is impossible [because, according to the bisection of angle HOT by HM, HM:MT = HO:TO, whereas, according to the bisection of angle HZT by ZL, HZ:TZ = HL:HT, so it should necessarily follow that HM:MT = HL:LT, because M and L should cut equiproportional segments from HT]. 
◉ Palam igitur quod, si T et H equaliter distant a centro et fuerint supra contingentem, non refertur T ad H nisi ab uno speculi puncto tantum, et unicus erit ei ymaginis locus. 
◉ It is therefore obvious that, if T and H lie the same distance from the center [of the mirror], and if they lie above the tangent, [the form of point] T is reflected to H from only one point on the mirror, and it will have a single imagelocation.⁑ 
◉ Amplius, BDQ, ADG [FIGURE 5.2.37b, p. 596] sint duo dyametri spere, et dyameter EDZ dividat angulum BDG per equalia, et a puncto E ducantur due perpendiculares super duos dyametros BD, GD, que sunt ET, EH. 
◉ Now, let BDQ and ADG [figure 5.2.37b, p. 258] be two diameters within the sphere [from which the mirror is formed], let diameter EDZ bisect angle BDG, and from point E draw the two [lines] ET and EH perpendicular to the two diameters BD and GD. 
◉ Palam quod triangulus ETD equalis est triangulo EHD, cum ED sit communis utrique. T igitur refertur ad H a puncto E. Eodem modo, a puncto Z. Et palam quod non refertur ad E ab aliquo puncto arcus AB vel arcus GQ, nec refertur ab alio puncto arcus AQ quam a puncto Z, secundum supradictam probationem. Verum quod ab alio puncto arcus BG quam a puncto E non possit reflecti patebit sic. 
◉ It is obvious that triangle ETD = triangle EHD, since ED is common to both [and angles EHD and ETD, as well as EDH and ETD are equal, by construction]. Thus, [the form of point] T is reflected to H from point E. By the same token, [it is reflected to point T] from point Z. It is also obvious [from proposition 34 above] that it is not reflected to [point] E from any point on arc AB or arc GQ, nor is it reflected from any point on arc AQ other than from point Z, according to the previous demonstration [in case 1 above]. But that it cannot be reflected from any point on arc BG other than from point E will be shown as follows. 
◉ Detur O punctum, et ducantur linee TO, HO DO. Fiat circulus ad quantitatem linee DE transiens per tria puncta T, D, H, cuius quidem circuli linea DE erit dyameter, cum angulus ETD quem respicit sit rectus. Igitur circulus ille transibit per punctum E. 
◉ Given point O [from which reflection supposedly does occur], draw lines TO, HO, and DO. Construct a circle the size of DE [as diameter] that passes through the three points T, D, and H, and line DE will be the diameter [by Euclid, III.31], since the angle ETD that it subtends is a right angle [by construction]. Therefore, that circle will pass through point E. 
◉ Cum igitur E sit communis utrique circulo, et sit super eundem dyametrum, continget circulus minor maiorem in puncto E, sicut probat Euclides. Igitur circulus iste secabit lineam DO. 
◉ Accordingly, since E is common to each circle [i.e., TDHE and AZGE], and since it lies on the same diameter [within each circle], the smaller circle will be tangent to the larger circle at point E, as Euclid demonstrates [in III.11]. Therefore, this [smaller] circle will intersect line DO. 
◉ Secet in puncto I, et ducantur linee TI, HI. Iam patet quod TD equalis est DH. Igitur angulus TID equalis angulo DIH, quia super equales arcus. Restat angulus TIO equalis angulo HIO; et angulus IOT est equalis angulo IOH, ex ypotesi, et IO commune. Erit triangulus TIO equalis triangulo HIO, et erit TO equalis HO, quod est impossibile, quoniam HO maior HE, TO minor TE, et TE, sicut prius probatum est, equalis est HE. Restat ergo ut T non reflectatur ad H ab alio puncto quam ab E vel a Z. 
◉ Let it intersect at point I, and draw lines TI and HI. It is already evident that TD = DH [since they are corresponding sides of equal triangles TED and HED]. Thus, angle TID = angle DIH, since they are subtended by equal arcs [DT and DH in circle HTE]. It follows that [being adjacent to angle TID] angle TIO = angle HIO [which is adjacent to angle DIH]; angle IOT = angle IOH, by construction, and IO is common. [Hence] triangle TIO = triangle HIO [by Euclid, I.26], and TO = HO, which is impossible, since [by Euclid, III.7] HO > HE, [while] TO < TE, and TE = HE, as has been proven earlier. It therefore follows that [the form of point] T may not be reflected to H from any point other than E or Z. 
◉ Iterum, a puncto E ducatur linea super dyametrum TD, qui sit EM, et secetur a linea HD pars equalis MD, que sit ND, et ducantur EN, EM. Palam quod angulus EMD est maior recto. Secetur ex eo equalis recto per lineam CM, que concurret cum DE. Sit concursus punctus C. et ducatur NC, et fiat circulus ad quantitatem CD transiens per tria puncta M, D, N. Cum CMD sit rectus, erit CD dyameter, et transibit circulus per C. Palam ergo quod M refertur ad N a puncto E, et similiter a puncto Z, et non ab aliquo puncto arcus AB vel QG; et palam quod non ab alio puncto arcus AQ quam a puncto Z. 
◉ Moreover, from point E draw line EM to diameter TD, from line HD cut off a segment ND equal to MD, and draw EN and EM. It is evident that angle EMD is greater than a right angle [because it is exterior to right angle ETM in triangle ETM]. Cut from it [an angle] equal to a right angle with line CM, which will intersect DE. Let C be the point of intersection, draw NC, and construct a circle according to the size [of diameter] CD that passes through the three points M, D, and N. Since CMD is a right angle [by construction], CD will be the diameter, and the circle will pass through C. It is therefore obvious that [the form of point] M is reflected to N from point E [because triangle NCE = triangle MCE, so angle NED = corresponding angle MED], and likewise [it is reflected] from point Z, but from no other point on arc AB or QG; and it is obvious that [it does not reflect] from any point on arc AQ other than point Z. 
◉ Et quod non ab alio puncto arcus BG quam a puncto E palam secundum modum predictum. Sumpto enim puncto, et ductis lineis ad punctum illud a punctis T, D, H, et sumpto puncto in quo circulus ultimus secabit dyametrum, et a puncto sectionis ductis lineis ad puncta T, H, eadem erit improbatio que prius. 
◉ That [it does not reflect] from any point on arc BG other than from point E is evident from the earlier procedure. For if [some such] point is taken, if lines are drawn to that point from points T, D, and H, if a point is taken where the last circle will cut the diameter, and if lines are drawn to points T and H from the point of intersection, the same disproof as before will apply. 
◉ Palam ergo ex predictis quod si angulum contentum a duobus dyametris per equalia dividat tertius dyameter, et a termino illius dyametri ducantur perpendiculares ad illos dyametros, puncta dyametrorum in que cadunt ad se invicem reflectuntur a duobus punctis speculi tantum. Puncta etiam dyametrorum citra hos terminos perpendicularium sumpta, id est, versus centrum, reflectitur quodlibet a duobus punctis tantum, et unum refertur ad illud quod equaliter distat a centro, et omnium talium duplex est ymaginis locus. 
◉ From the foregoing, then, it is clear that, if a third diameter bisects the angle formed by two [other] diameters, and if from the endpoint of that [third] diameter perpendiculars are drawn to those [other two] diameters, the points on the diameters where those perpendiculars fall are reflected to one another from only two points on the mirror. In addition, any of the points taken on the perpendiculars to the diameters below these endpoints, i.e., toward the center, is reflected from only two points [on the circle], and the one is reflected to the other that lies equidistant from the center, and for all such points the imagelocation is twofold.⁑ 
◉ Amplius, sumptis duobus dyametris BQ, AG, et EZ [FIGURE 5.2.37c, p. 597] dividente angulum eorum per equalia, et sumatur in BD punctus T supra punctum in quem cadit perpendicularis ducta a puncto E, et in DG sumatur DH equalis DT, et ducantur TE, HE. Refertur quidem T ad H a puncto E, et similiter a puncto Z, non ab alio puncto arcus AQ, nec ab aliquo puncto arcus AB vel GQ. 
◉ Furthermore, given the two diameters BQ and AG [figure 5.2.37c, p. 259], and given that EZ bisects the angle formed by them, take point T on BD beyond the point [between D and T] where the perpendicular dropped from point E falls [upon DB], take DH on DG equal to DT, and draw TE and HE. [Thus, according to the preceding case, the form of point] T is reflected to H from point E, and likewise from point Z, [but] not from any other point on arc AQ, or from any point on arcs AB or GQ. 
◉ Deinceps, a puncto T ducatur perpendicularis super TD, que quidem concurret cum DE extra circulum spere, cum angulus DTE sit acutus. Concurrat ergo in puncto O, et ducantur linee TO, HO. Et fiat circulus transiens per tria puncta T, D, H, qui necessario transibit per punctum O, et erit DO dyameter eius. Et ducantur linee TO, HO, et ducatur linea contingens circulum BZG in puncto E, que sit KE. Palam quoniam ultimus circulus secabit primum, scilicet BZG, in duobus punctis. Sint puncta illa L, M, et ducantur linee TL, HL, LD, TM, DM, HM. [2.369] Cum ergo arcus TD sit equalis arcui HD, erit angulus TLD equalis angulo DLH, et ita T refertur ad H a puncto L. Similiter, angulus TMD equalis angulo DMH, et ita T refertur ad H a puncto M. Palam igitur quod T refertur ad H a quatuor punctis, scilicet E, Z, L, M, et quadruplex erit locus ymaginis eius. 
◉ Then, from point T on TD draw a perpendicular, which will intersect DE outside the circle on the [mirror’s defining] sphere, because angle DTE is acute. Accordingly, let it intersect at point O, and draw lines TO and HO. Construct a circle passing through the three points T, D, and H. That circle will necessarily pass through point O, and DO will be its diameter. Draw lines TO and HO, and draw line KE tangent to circle BZG at point E. It is clear that the latter circle will intersect the first one, i.e., BZG, at two points. Let those points be L and M, and draw lines TL, HL, LD, TM, DM, and HM. 
◉ Et non potest T reflecti ad H ab alio puncto quam aliquo istorum. Detur enim punctus F, et ducantur linee TF, HF, DF, et producatur DF usque concurrat cum contingenti KE, et sit concursus K. Et ducantur linee TK, HK. Igitur angulus TFD equalis angulo DFH, ex ypotesi; restat angulus TFK equalis angulo KFH. Sed angulus TKF equalis est angulo FKH, quia super equales arcus, et FK communis. Erit triangulus equalis triangulo, et ita TK equalis KH, quod est impossibile, quoniam HK maior HO, et TK minor TO, et TO equalis HO. 
◉ But [the form of point] T cannot be reflected to H from any point other than these. For, let F be given [as such a point], draw lines TF, HF, and DF, extend DF until it intersects tangent KE, and let K be the intersection. Then draw lines TK and HK. Hence, angle TFD = angle DFH, by construction; and it follows that [being adjacent to angle TFD] angle TFK = angle KFH [which is adjacent to angle DFH]. But angle TKF[D] = angle [D]FKH, because they are subtended by equal arcs, and FK is common. [Therefore] triangle [TKF] = triangle [HFK], and so TK = KH, which is impossible, because HK > HO [since arc HK > arc HO, whereas] TK < TO, yet TO = HO [by construction]. 
◉ Palam igitur quod non fit reflexio ab aliquo puncto quam a punctis quatuor circuli. 
◉ It is evident, then, that reflection does not occur from any point other than from the four [previously defined] points on the circle. 
◉ Igitur, si in diversis dyametris sumantur duo puncta, scilicet T, H, equaliter a centro distantia, si fuerint super puncta dyametrorum in que cadunt perpendiculares ducte a termino diametri dividentis per equalia angulum duorum dyametrorum, aut fuerint inter centrum et puncta illa, id est citra perpendiculares, dum equaliter distent a centro, reflectetur quidem T ad H a duobus punctis tantum. 
◉ [In summary] therefore, if two points, i.e., T and H, are taken on different diameters [at locations on those diameters that are] equidistant from the center, and if they lie at the points on the diameters where the perpendiculars drawn from the end of the diameter that bisects the angle formed by those two diameters fall, or if they lie between the center and those points, i.e., below the perpendiculars, as long as they lie equidistant from the center, then [the form of point] T will be reflected to H from two points only [as demonstrated in case 2 above]. 
◉ Si vero T et H fuerint a locis perpendicularium usque in circulum, reflectetur quidem T ad H a quatuor punctis. Si vero fuerint in circulo vel extra, tamen citra contingentem KE, reflectetur quidem T ad H a duobus punctis tantum. Si vero supra contingentem fuerint, reflectetur quidem T ad H ab uno puncto tantum. Et hec quidem accidunt, dum T equaliter distet a centro cum puncto H. 
◉ If, however, T and H lie at spots beyond the perpendiculars up to the interior surface of the circle, [the form of point] T will be reflected to [point] H from four points. But if they lie on the circle or outside it, but still below tangent KE, [the form of point] T will be reflected to H from two points only [i.e., E and Z]. And if they lie upon the tangent, [the form of point] T will be reflected to H from only one point [i.e., Z].⁑ And these phenomena occur as long as T lies the same distance as point H from the center. 
◉ [PROPOSITIO 38] Amplius, si fuerint T, H in diversis dyametris, et longitudo eorum a centro fuerit inequalis, reflectentur quidem ad se ab uno puncto. 
◉ [PROPOSITION 38] Now, if T and H lie on different diameters, and if they lie unequal distances from the center, they will be reflected to one another from one point [only on the opposite arc].⁑ 
◉ Verbi gratia, ducantur dyametri ADG, BDQ [FIGURE 5.2.38, p. 598], et EZ dividat angulum eorum per equalia. Et T propinquior sit centro D quam H. Et sumatur linea LQ, et dividatur in puncto M ut sit proportio QM ad ML sicut HD ad DT. Et dividatur LQ per equalia in punctum N, et a puncto N ducatur perpendicularis NK, et super punctum L fiat angulus equalis medietati anguli ADT per lineam FL. Erit quidem angulus FLQ acutus, quare FL concurret cum NK. Concurrat in puncto F, et a puncto M ducatur linea ad latus FL concurrens cum latere NK in puncto quod sit K. Et secet linea illa latus FL in puncto C ut sit proportio KC ad CL sicut HD ad DZ. 
◉ For instance, draw diameters ADG and BDQ’ [figure 5.2.38, p. 260], and let EZ bisect the angle formed by them. Let [objectpoint] T be nearer centerpoint D than [center of sight] H. Take [some arbitrary] line LQ and cut it at point M so that QM:ML = HD:DT [by Euclid, VI.10]. Bisect LQ at point N, from point N draw perpendicular NK, and at point L form with line FL an angle [FLQ] equal to half of angle ADT. Angle FLQ will be acute, so FL will intersect NK. Let it intersect at point F, and from point M draw a line to side FL intersecting side NK at point K. Let that line cut side FL at point C such that KC:CL = HD:DZ [by proposition 21, lemma 3 above]. 
◉ Deinceps super punctum D fiat angulus equalis angulo LCM, qui sit IDA, et sit I punctus circuli supra Z, aut infra. Et supra I punctum fiat angulus equalis CLM, qui sit OID, et super hanc lineam OI ducatur perpendicularis a puncto H, que sit HC, et producatur linea CF equalis linee CI, et ducantur linee HF, HI. 
◉ Then, on point D form angle IDA equal to angle LCM, and let I be a point on the circle lying above or below Z. At point I form angle O’ID equal to angle CLM, to the resulting line O’I drop perpendicular HC’ from point H, draw line C’F’ equal to line C’I, and draw lines HF’ and HI. 
◉ Palam secundum predicta quod a puncto M non potest linea duci ad latus FL dividens ipsum eo modo quo dividit linea MCK preter hanc solam lineam MCK. Si enim possit, sit MPO. Palam quoniam PO minor erit CK, quod quidem patebit, ducta linea PY equidistans CK, que erit minor CK et maior PO. Et PL maior CL. Igitur non erit proportio PO ad PL sicut KC ad CL, quare non erit PO ad PL sicut HD ad DT. Restat ergo ut a puncto M non ducatur alia quam MCK similis ei. 
◉ According to previous arguments, it is clear that from point M no line other than line MCK can be drawn to side FL to cut it in the way that line MCK does. For if [such a line] could [be found], let it be MPO. It is evident that PO < CK, which will become evident if line PY is drawn parallel to CK, that line being shorter than CK and longer than PO. But PL > CL. Thus, PO:PL ≠ KC:CL, so PO:PL ≠ HD:DT. It therefore follows that no line like MCK, other than MCK [itself], can be drawn from point M [to side FL]. 
◉ Verum cum ODI sit equalis angulo LCM, et angulus OID equalis angulo CLM, erit triangulus CLM similis triangulo IOD. Igitur angulus IOD erit equalis angulo LMC. Restat angulus COH equalis angulo KMN, et angulus HCO rectus equalis angulo KNM. Restat angulus NKM equalis angulo CHO. 
◉ Now, since [angle] O’DI = angle LCM [by construction], and angle O’ID = angle CLM [by construction], triangle CLM will be similar to triangle IO’D. Therefore angle IO’D = [corresponding] angle LMC. It follows that [being adjacent to angle IO’D] angle C’O’H = angle KMN [which is adjacent to angle LMC], and right angle HC’O’ = [right] angle KNM [both being right angles by construction]. [So] it follows that angle NKM [in triangle NKM] = angle C’HO’ [in triangle C’HO’, so triangles NKM and C’HO’ are similar, by Euclid, VI.4]. 
◉ Ducta autem linea DI donec concurrat cum HC in puncto R, erit angulus RDH equalis angulo KCF. Erit triangulus RDH similis triangulo CKF. Igitur proportio RD ad DH sicut FC ad KC. Sed HD ad DI sicut KC ad CL. Igitur RD ad DI sicut FC ad CL. Igitur RI ad DI sicut FL ad CL. Sed DI ad IO sicut CL ad LM, cum triangulus DIO sit similis triangulo CLM. Igitur RI ad IO sicut FL ad LM. Sed proportio RI ad IC sicut FL ad LN, quoniam triangulus RIC similis est triangulo FLN. Igitur, proportio IO ad IC sicut LM ad LN. Igitur proportio QM ad LM sicut FO ad IO. 
◉ Now, if line DI is extended until it intersects HC’ at point R, angle RDH = angle KCF [because angle KCF = vertical angle LCM, angle RDH = vertical angle O’DI , and angle LCM = angle O’DI, by construction]. Triangle RDH will [thus] be similar to triangle CKF [since corresponding angles RHD and CKF are also equal, by previous conclusions]. Therefore, RD:DH = FC:KC. But HD:DI = KC:CL [by construction]. So RD:DI = FC:CL [by Euclid, V.22]. Accordingly, RI:DI = FL:CL [by Euclid, V.18]. But DI:IO’ = CL:LM, since triangle DIO’ is similar to triangle CLM [by previous conclusions]. Therefore, RI:IO’ = FL:LM [by Euclid, V.22]. But RI:IC’ = FL:LN, because triangle RIC’ is similar to triangle FLN [since right angle RC’I = right angle LNF, and angle RIC’ = corresponding angle FLN]. Hence, IO’:IC’ = LM:LN [by Euclid, V.22]. So QM:LM = F’O’:IO’.⁑ 
◉ Ducta autem a puncto I linea UI equidistante linee HF, et producta linea DA donec concurrat cum UI in puncto U, erit triangulus OUI triangulo HOF similis. Erit igitur proportio HO ad OU sicut QM ad ML, et ita HO ad OU sicut HD ad DT. Sed quoniam triangulus HCI equalis est triangulo HCF, cum HC sit perpendicularis, igitur angulus HFC equalis est angulo CIH, et ita CIH equalis est angulo UIO, quare proportio HO ad OU sicut HI ad UI, et ita HI ad UI sicut HD ad DT. 
◉ Now, if line UI is drawn from point I parallel to HF’, and if line DA is extended until it intersects UI at point U, triangle O’UI will be similar to triangle HO’F’. Therefore, HO’:O’U = QM:ML [because HO’:O’U = F’O’:O’I], and so HO’:O’U = HD:DT [because QM:ML = HD:DT, by construction]. But since triangle HC’I = triangle HC’F’, because HC’ is perpendicular [to equal bases F’C’ and C’I], then angle HF’C’ = angle C’IH, so [angle] C’IH = angle UIO’, and so [by Euclid, VI.3] HO’:O’U = HI:UI [because angle UIH in triangle UIH is bisected by IO’, insofar as angle UHI is composed of equal angles UIO’ and C’IH], and so HI:UI = HD:DT. 
◉ Verum angulus UID maior est angulo DIH. Secetur ab eo equalis, et sit PID, et ducatur linea TP. P sit punctus dyametri DA. 
◉ But angle UID > angle DIH [since angle UID > angle UIO’, which = angle O’IH, of which DIH is a part]. From it [i.e., angle UID] cut angle PID equal [to DIH], and draw line TP. Let P be a point on diameter DA. 
◉ Palam quod proportio HI ad UI constat ex proportione HI ad IP et IP ad UI, et proportio HI ad IP sicut HD ad DP, quoniam DI dividit angulum PIH per equalia. Igitur proportio HI ad UI, que est HD ad DT, constat ex proportione HD ad DP et PI ad UI. Sed proportio HD ad DT constat ex proportione HD ad DP et DP ad DT. Igitur proportio DP ad DT sicut PI ad UI. 
◉ It is evident that the ratio HI:UI is compounded from HI:IP and IP:UI, while HI:IP = HD:DP [by Euclid, VI.3], because DI bisects angle PIH [in triangle PIH]. Therefore, HI:UI, which is as HD:DP, is compounded of the ratios HD:DP [which = HI:IP] and PI:UI. But the ratio HD:DT is compounded of HD:DP and DP:DT. Thus, DP:DT = PI:UI.⁑ 
◉ Verum angulus OIH est medietas anguli UIH, sed angulus DIH est medietas anguli PIH. Restat angulus DIO medietas anguli PIU, sed angulus DIO est medietas anguli TDP, quia est equalis angulo FLM. Igitur angulus PIU est equalis angulo TDP, et proportio DP ad DT sicut PI ad UI. Igitur triangulus UIP similis est triangulo TPD, et angulus UPI equalis TPD. Erit igitur TPI linea recta, quia angulus DPT cum angulo TPO valet duos rectos, et ita angulus OPI cum angulo OPT valet duos rectos. Et ita T refertur ad H a puncto I, et hec quidem erit probatio sive sit T extra circulum, sive intra, et similiter, sumpto puncto H extra vel intra, dum inequaliter a centro. 
◉ Now, angle O’IH is half of angle UIH [by previous conclusions], whereas angle DIH is half of angle PIH [by construction]. It follows that angle DIO’ is half of angle PIU, whereas angle DIO’ is half of angle TDP, since it is equal to angle FLM. Therefore, angle PIU = angle TDP,⁑ and DP:DT = PI:UI [by previous conclusions]. Hence, triangle UIP is similar to triangle TPD [by Euclid, VI.4], and [so] angle UPI = [angle] TPD. TPI will thus be a straight line, because angle DPT + angle TPO’ = two right angles, and so angle O’PI + angle O’PT = two right angles [since O’PI is alternate to DPT]. So [the form of point] T is reflected to [point] H from point I, and this proof will apply whether T lies outside or inside the circle, and [it will apply] likewise if point H is taken outside or inside [the circle], as long as [they are] unequally distant from the center.⁑ 
◉ [PROPOSITIO 39] Amplius, ductis dyametris BQ, AG, et dyametro EZ dividente angulum BDG per equalia, dico quoniam, quicumque punctus sumatur in arcu AQ preter punctum Z, ab illo poterunt reflecti infinita paria punctorum inequaliter a centro distantium. 
◉ [PROPOSITION 39] Now, having produced diameters BQ and AG, and [given] diameter EZ bisecting angle BDG, I say that, no matter what point other than Z is taken on arc AQ, an infinite number of pointpairs that are not equidistant from the center can be reflected [to one another] from that point. 
◉ Verbi gratia, sumatur punctus H [FIGURE 5.2.39, p. 599], et sumatur in dyametro GD punctus L. Et a dyametro BD secetur MD equalis LD, et ducantur linee LM, LH, MH, DH. Punctus in quo EZ dividit LM sit F; erit LF equalis FM. 
◉ For example, take point H [as the potential point of reflection in figure 5.2.39, p. 261], and take point L on diameter GD. On diameter BD cut off [segment] MD equal to LD, and draw lines LM, LH, MH, and DH. Let F be the point where EZ cuts LM; [and so] LF = FM [since triangles FMD and LDF are equal, by Euclid, I.4]. 
◉ Et ducatur HD usque cadat super LM in puncto N. Erit igitur LN minor NM. Verum, cum angulus MDF sit equalis FDL et angulo QDZ, et angulus MDA equalis angulo LDQ, et angulus ADH equalis angulo NDL, erit angulus LDH maior angulo MDH. Igitur LH erit maior MH, cum MD, DH equalia sint LD, DH. Erit ergo angulus DHL minor angulo DHM, si enim esset equalis, esset proportio LH ad MH sicut LN ad NM, quod est impossibile. Si autem fuerit maior, secetur ex eo equalis, et improbetur hoc modo. Igitur est minor. 
◉ Draw [line] HD until it falls on LM at point N. Hence, LN < MN. But, since angle MDF = [angle] FDL, as well as [vertical] angle QDZ, since angle MDA = [vertical] angle LDQ, and since angle ADH = [vertical] angle NDL, then angle LDH > angle MDH.⁑ Therefore [by Euclid, I.24], LH > MH, since MD and DH = LD and DH [respectively]. Thus, angle DHL < angle DHM, for if they were equal [then DHN would bisect angle MHD, and so, by Euclid, VI.3], LH:MH would be as LN:NM, which is impossible. If, on the other hand, it were greater [i.e., if DHL > DHM], then cut from it [an angle] equal [to DHM], and it will be disproven in this [same] way.⁑ Therefore it is smaller [i.e., DHL < DHM]. 
◉ Secetur igitur ab angulo MHD equalis illi, qui sit THD. Igitur T refertur ad L a puncto H, et TD minor LD. 
◉ Accordingly from angle MHD cut off [angle] THD equal to it [i.e., DHL]. [The form of point] T is therefore reflected to [point] L from point H, and TD < LD [i.e., T and L lie unequal distances from D]. 
◉ Similiter, si sumantur in dyametris HD, GD alia puncta quam L, M equaliter a puncto D distantia, probatur similiter quod a puncto H fit reflexio punctorum adinvicem et inequaliter distantium a centro. Et ita de infinitis punctis in hiis dyametris sumptis similis erit probatio, et a quocumque puncto arcus AQ sumpto preter quam a puncto Z. 
◉ So, too, if points other than L and M that lie equidistant from point D are selected on diameters HD and GD, it is proven in a similar way that from point H there occurs a mutual reflection of points that lie at different distances from the center. And so the proof will be the same for an infinite number of points chosen on these diameters, as well as for any point chosen on arc AQ other than point Z [because, in order for reflection to occur from Z, the points must be equidistant from the center, as shown in proposition 37].⁑ 
◉ [PROPOSITIO 40] Amplius, sumptis T, L [FIGURE 5.2.40, p. 599] in dyametris quorum inequalis sit longitudo a centro, et reflectantur adinvicem a puncto H, non erit reflecti T ad L ab alio puncto arcus AQ quam a puncto H. 
◉ [PROPOSITION 40] Moreover, if T and L [figure 5.2.40, p. 261] are taken on [different] diameters at unequal distances from the center, and if they are reflected to one another from point H, [the form of point] T will not be reflected to [point] L from any point on arc AQ other than from point H. 
◉ Si enim ab alio, sit illud K, et ducantur TK, LK, DK, LT, TH, LH, NDH. Et producatur DK usque cadat in LT in puncto C. Palam quoniam proportio LH ad TH sicut LN ad NT. 
◉ For if [it were reflected] from another [point], let it be K, and draw TK, LK, DK, LT, TH, LH, and NDH. Then extend DK until it falls to point C on LT. It is clear that LH:TH = LN:NT [by Euclid, VI.3]. 
◉ Et similiter, cum angulus TKC sit equalis angulo LKC, ex ypotesi, erit proportio LK ad TK sicut LC ad CT. Sed LH maior LK, et TH minor TK. Igitur maior est proportio LH ad TH quam LK ad TK, quare maior erit proportio LN ad NT quam LC ad CT, quod plane impossibile. Restat ut ab alio puncto arcus AQ quam a puncto H non possit T reflecti ad L. Palam ergo que accidunt in arcu AQ. 
◉ Likewise, since angle TKC = angle LKC, by supposition, then LK:TK = LC:CT. But [by Euclid, III.7] LH > LK, and TH < TK. Hence LH:TH > LK:TK, so LN:NT > LC:CT, which is patently impossible. It follows that [the form of point] T cannot be reflected to L from any point on arc AQ other than from point H. What pertains to arc AQ is therefore evident. 
◉ [PROPOSITIO 41] Amplius, sit A [FIGURE 5.2.41, p. 600] centrum visus, B centrum speculi, et ducatur dyameter DABG. Et sumatur superficies in qua sit AB quocumque modo que secabit speram super circulum qui sit DLG. Dico quod a quolibet puncto semicirculi DLG reflectuntur puncta ad A inequalis longitudinis a centro cum eo. 
◉ [PROPOSITION 41] To continue, let A [figure 5.2.41, p. 262] be the center of sight, B the center of the mirror, and draw diameter DABG. Take the plane in which AB lies so that it will somehow intersect the sphere along [great] circle DLG. I say that points that do not lie the same distance from the center as A are reflected [to A] from any point on semicircle DLG. 
◉ Verbi gratia, sumatur punctus E, et ducantur linee EA, EB. Palam quoniam angulus AEB erit acutus, quia cadet in minorem arcum semicirculo. Fiat ei equalis, et sit OEB, et ducatur linea OE quantumlibet. Palam quod quodlibet punctum illius linee refertur ad A a puncto E. 
◉ For instance, take point E, and draw lines EA and EB. It is obvious that angle AEB will be acute, because it will be subtended by a smaller arc than a semicircle. Form [angle] OEB equal to it, and draw line OE as far as you like. It is evident that [the form of] every point on that line is reflected to A from point E. 
◉ Ducta autem a puncto B ad lineam OE perpendiculari, aut erit perpendicularis illa equalis BA, aut maior, aut minor. Si fuerit equalis, linee omnes ducte a puncto B ad lineam OE, preter illam perpendicularem, erunt maiores linea BA, et ita quodlibet punctum linee OE, uno excepto, inequaliter distabit a centro puncto A. 
◉ Now, if a perpendicular is dropped from point B to line OE, that perpendicular will be equal to, longer than, or shorter than BA. If it is equal, then every line, other than the perpendicular, that is drawn from point B to line OE will be longer than line BA, and so, with one exception [i.e., point O], every point on line OE will lie a different distance from the center than point A. 
◉ Si vero perpendicularis fuerit maior, omnia puncta linee illius plus distabunt a centro quam A punctum. Si autem perpendicularis fuerit minor, erit ducere a puncto B duas lineas ex diversis partibus perpendicularis equales linee BA, et omnes alie linee aut minores erunt aut maiores. Palam igitur quoniam a puncto E reflectuntur puncta ad A quorum longitudo a centro inequalis est longitudini A ab eodem, quod est propositum. 
◉ If the perpendicular is longer [than AB], then all the points on that line [i.e., OE] will lie farther from the center than point A [and will therefore lie different distances from B than A]. But if the perpendicular is shorter, two lines equal to BA can be drawn from [points on] opposite sides of the perpendicular, whereas all the remaining lines will be either shorter or longer [than BA and will therefore lie at different distances from B]. It is therefore evident that [the forms of] points that do not lie the same distance from the center as [point] A are reflected to [point] A from point E, which is what was proposed. 
◉ Constat ex hiis quod, si sumatur A extra circulum—et sit H [FIGURE 5.41a, p. 600]—et ducatur dyameter HDBG et due contingentes HT, HQ, a quolibet puncto arcus TG, preter quam a T vel G, poterit fieri reflexio ad H punctorum inequaliter distantium a centro cum puncto H. Et erit eadem probatio. 
◉ From these considerations it is clear that, if [center of sight] A is taken outside the circle—let it be at H [figure 5.2.41a, p. 262]—and if diameter HDBG is drawn along with tangents HT and HQ, then reflection of points that do not lie the same distance from the center as H can take place to H from any point on arc TG other than from T or G. And the proof will be the same [as the previous one].⁑ 
◉ [PROPOSITIO 42] Amplius, ex hiis constabit quod, facta reflexione ad A a puncto E vel alio puncto inequaliter distante a centro puncto A, dyameter in quo fuerit punctus reflexus cum dyametro ABG facit duos angulos, unum respicientem angulum reflexionis, alium ei collateralem, qui quidem collateralis aliquando erit maior angulo reflexionis, aliquando minor. [2.398] Verbi gratia, ducatur perpendicularis FB [FIGURE 5.2.42, p. 600] super EO. BA aut est perpendicularis super ea aut non. 
◉ [PROPOSITION 42] On the basis of these determinations, it will be established that, if reflection occurs to [point] A from point E or from another point not lying the same distance as point A from the center, the diameter in which the point [whose form is] reflected lies forms two angles with diameter ABG, one of them opposite the reflected angle, the other adjacent to it, and that adjacent angle will sometimes be greater than the reflected angle and sometimes smaller.⁑ 
◉ Sit perpendicularis. Erit igitur EA equidistans FB, et erunt duo anguli FBA, FEA equales duobus rectis. Ducta autem linea BO, erunt duo anguli OBA, OEA minores duobus rectis.. Igitur erit angulus OBG maior angulo OEA, qui est angulus reflexionis. Et cum triangulus EBF sit equalis triangulo EBA, erit BF equalis BA, et ita OB maior BA. 
◉ Let it be perpendicular. Accordingly, EA will be parallel to FB, and the two angles FBA and FEA will sum up to two right angles. Now, when line BO is drawn, the two angles OBA and OEA will sum up to less than two right angles. Hence, angle OBG > angle OEA, which is the reflected angle. And since triangle EBF = triangle EBA, BF = BA, and so OB > BA. 
◉ Ducta autem linea BN, erunt duo anguli NBA, NEA maiores duobus rectis. Erit ergo angulus NBG minor angulo NEA, et NB maior BA, et ita N et O reflectuntur ad A a puncto E. Et inequaliter distant a centro puncto A, et dyameter OB cum dyametro ABG ex parte G facit angulum maiorem angulo reflexionis, et dyameter NB maiorem, et ita propositum. 
◉ Moreover, when line BN is drawn, the two angles NBA and NEA will sum up to more than two right angles. Thus, angle NBG < angle NEA, and NB > BA, and so [the forms of points] N and O will be reflected to [point] A from point E. Also, they lie at different distances than point A from the center, and diameter OB forms an angle with ABG that is greater on the side of G than the reflected angle [OEA], while diameter NB is longer [than diameter AB], so [we have demonstrated] what was proposed. 
◉ Si vero BA non fuerit perpendicularis super EA, ducatur perpendicularis, que sit BK, que quidem sive cadat supra AB [FIGURE 5.2.42a, p. 600], aut sub [FIGURE 5.2.42b, p. 600]. Eadem erit probatio. 
◉ If, however, BA is not perpendicular to EA, then draw the perpendicular BK, which either lies above AB [figure 5.2.42a, p. 262], or lies below it [figure 5.2.42b, p. 262]. The proof will be the same [in both cases]. 
◉ Et BF sit perpendicularis super EO, et ducatur FT equalis AK, et ducatur TB. Palam quoniam in triangulo KEB angulus EKB rectus equalis EFB, et angulus KEB equalis angulo FEB. Restat tertius tertio equalis, et cum EB sit latus commune utrique triangulo, erunt trianguli equales, et erit FB equalis KB. Sed AK est equalis FT. Erit AB equalis BT, et angulus ABK equalis angulo FBT. 
◉ Let BF be perpendicular to EO, draw FT equal to AK, and draw TB. It is evident that in triangle KEB right angle EKB = [right angle] EFB, and angle KEB = angle FEB [by construction]. It follows that the third [angle, i.e., EBK] = the third [angle, i.e., EBF], and since side EB is common to both triangles [EBF and EBK], the triangles will be equal, and FB = KB. But AK = FT [by construction. And since triangle BAK = triangle BTF] AB = BT, and angle ABK = angle FBT. 
◉ Addito communi angulo FBA, erit KBF equalis TBA. Sed KBF et FEA valent duos rectos, quare TBA, TEA valent duos rectos, et ita TBG equalis est angulo TEA, qui est angulus reflexionis. 
◉ With common angle [KBT] added to [both in order to yield angle] FBA, [angle] KBF = [angle] TBA.⁑ But [angles] KBF and FEA sum up to two right angles,⁑ so [angles] TBA and TEA sum up to two right angles, and so [angle] TBG = angle TEA, which is the reflected angle. 
◉ Si igitur a puncto B ad lineam ET ducatur linea ultra T, faciet angulum cum BG ex parte G minorem angulo reflexionis. Et erit linea illa maior AB, quoniam TB equalis AB. 
◉ Therefore, if a line is drawn from point B to [fall on] line ET beyond T, it will form an angle with BG on the side of G that is smaller than the reflected angle. And that line will be longer than AB, because TB = AB. 
◉ Et quelibet linea a puncto B ducta ad ET et citra T faciet angulum cum BG ex parte G maiorem angulo reflexionis, et erit inequalis AB, et ita propositum. 
◉ Furthermore, any line drawn from point B to [fall on] line ET in front of T [i.e., between T and E] will form an angle with BG that is greater on the side of G than the reflected angle, and it will be unequal to AB, so [we have demonstrated] what was proposed.⁑ 
◉ [PROPOSITIO 43] Amplius, sit B centrum visus, G centrum spere. Ducatur dyameter ZBGD [FIGURE 5.2.43, p. 601], et sumatur superficies in qua sit dyameter secans speram super circulum ZEH. Dico quoniam, si punctus A refertur ad B ab aliquo puncto circuli, et inequalis est distantia puncti A a centro et puncti B ab eodem, dyameter AG cum dyametro GD ex parte D faciet angulum quem impossibile est esse equalem angulo reflexionis. 
◉ [PROPOSITION 43] Now, let B be the center of sight and G the center of the sphere. Draw diameter ZBGD [figure 5.2.43, p. 263], and produce the plane that contains the diameter and cuts the sphere along [great] circle ZEH. I say that, if point A is reflected to point B from some point on the circle, and if the distance from point A to the center is not the same as that of point B from the center, diameter AG will form an angle [AGD] with diameter GD on the side of D that cannot possibly be equal to the reflected angle. 
◉ Sit equalis, et T sit punctus reflexionis, et sit AG inequalis BG. Ducantur linee TA, TG, TB, et fiat circulus transiens per tria puncta A, G, B, qui necessario transibit per punctum T. Si enim extra, ductis lineis a punctis A, B, ad idem punctum illius circuli extra fiet angulus minor angulo ATB. Et probabitur esse equalis. 
◉ Let it be equal, let T be the point of reflection, and let AG ≠ BG. Draw lines TA, TG, and TB, and construct a circle passing through the three points A, G, and B, and this circle will necessarily pass through point T. For if the circle lies beyond it, and if lines are drawn from points A and B to the corresponding point on that outlying circle, they will form an angle smaller than angle ATB. And it will be proven that it is equal. 
◉ Quoniam, cum angulo AGB valebit duos rectos, et angulus ATB, cum sit equalis angulo AGD, ex ypotesi, cum angulo AGB valet duos rectos, et ita impossibile. Similiter, si circulus citra T ceciderit, eadem erit improbatio. 
◉ For [angle AGD] along with angle AGB will sum up to two right angles, and angle ATB along with angle AGB sums up to two right angles, since angle ATB = angle AGD, by construction [and within the circle angle ATB + angle AGB = two right angles by Euclid, III.22], and so it is impossible [for T to lie outside the circle]. Likewise, if the circle were to pass below T, the same disproof would hold [insofar as the new angle ATB would be greater than ATB within the circle]. 
◉ Restat ergo ut transeat per punctum T, et cum angulus ATG sit equalis angulo BTG, erit arcus AG equalis arcui BG, et ita AG erit equalis BG. Et positum est esse inequalem, et ita propositum. 
◉ It follows from this that it must pass through point T, but since angle ATG = angle BTG [by supposition], arc AG = arc BG, and so [chord] AG = [chord] BG. Yet they have been assumed to be unequal, and so [we have demonstrated] what was proposed [i.e., that angle ATB cannot equal angle AGD when AG ≠ BG].⁑ 
◉ [PROPOSITIO 44] Amplius, sumptis in duobus dyametris EGH, ZGD [FIGURE 5.2.44, p. 602] duobus punctis A, B ut BG sit maior AG, dico quoniam, si punctus A refertur ad B a duobus punctis arcus EZ, non erit uterque angulus reflexionis minor angulo AGD. 
◉ [PROPOSITION 44] Furthermore, if the two points A and B [figure 5.2.44, p. 264] are taken on the two diameters EGH and ZGD so that BG > AG, I say that, if [the form of] point A is reflected to [point] B from two points on arc EZ, the angles of reflection will not both be smaller than angle AGD. 
◉ Sumantur enim duo puncta T, Q in arcu EZ a quibus A refertur ad B, scilicet T, Q, et ducantur linee BT, GT, AT, BQ, GQ, AQ. Et si angulus ATB minor est angulo AGD, dico quoniam angulus AQB non erit minor angulo AGD. 
◉ For, on arc EZ select two points, T and Q, from which [the form of point] A is reflected to [point] B, and draw lines BT, GT, AT, BQ, GQ, and AQ. If angle ATB < angle AGD, I say that angle AQB will not be smaller than angle AGD. 
◉ Sit enim minor, et ducatur linea GN dividens angulum dyametrorum per equalia, et ducatur linea AB quam dividat GN per punctum F. Palam quod proportio BG ad GA sicut BF ad FA. Sed BG maior GA; erit BF maior FA. 
◉ Let it in fact be smaller, draw line GN bisecting the angle formed by the diameters [i.e., angle BGA], and draw line AB, which GN intersects at point F. It is evident that BG:GA = BF:FA [by Euclid, VI.3]. But BG > GA; [so] BF > FA. 
◉ Dividatur AB per medium in puncto K, et fiat circulus transiens per tria puncta A, B, T, qui quidem circulus non transibit per G, quoniam essent anguli AGB, BTA equales duobus rectis, et palam quod sunt minores, cum angulus BTA sit minor angulo AGD. Igitur transibit supra G. 
◉ Bisect AB at point K, and construct a circle passing through the three points A, B, and T, a circle that will not pass through G, because [if it did] angles AGB and BTA [in quadrilateral BGTA] would be equal to two right angles [by Euclid, III.22], and it is obvious that they are less [than two right angles], since angle BTA < angle AGD [by supposition, and angle AGD + adjacent angle AGB = two right angles]. It will therefore pass above G. 
◉ Similiter, non transibit per Q. Quoniam, sumpto puncto circuli in quo linea GQ secat ipsum, scilicet M, esset arcus AM equalis arcui MB, cum respiciant equales angulos supra Q, quod manet impossibile, quoniam, sumpto O puncto in quo linea GT secat hunc circulum, erit arcus AO equalis arcui OB, quia respiciunt equales angulos supra T. Restat ut hic circulus transeat supra Q, si enim infra, eadem erit improbatio. 
◉ By the same token, it will not pass through Q. For if the point, i.e., M, is taken on the circle where line GQ intersects it, arc AM would be equal to arc MB, since [by Euclid, III.26] they would subtend equal angles [MQA and MQB] at Q, which is impossible, because, if point O, where line GT intersects this circle, is taken, arc AO = arc OB, since they subtend equal angles at T [and so GM, which supposedly bisects circle BMA, would cut a smaller arc on the side of A than on the side of B]. It follows that this circle passes above Q, for if it passed below, the same disproof would apply. 
◉ Ducatur autem linea a puncto O ad punctum K, que quidem, cum dividat cordam AB per equalia, et similiter arcum AB, erit perpendicularis super AB. Verum angulus BAG maior angulo ABG, cum BG maior GA. Et angulus BFG valet duos angulos FAG, FGA, et angulus AFG valet duos angulos FBG, FGB. 
◉ Now, draw a line from point O to point K [which is the midpoint of BA, by construction], and since it bisects chord AB and likewise bisects arc AB, this line will be perpendicular to AB. But [by Euclid, I.18] angle BAG > angle ABG, since BG > GA. Also, [exterior] angle BFG [of triangle GFA] equals the sum of the two [interior] angles FAG and FGA [by Euclid, I.32], while [exterior] angle AFG [of triangle BFG] equals the sum of the two [interior] angles FBG and FGB [by Euclid, I.32]. 
◉ Sed AGF equalis FGB, et FAG maior FBG. Igitur angulus BFG maior est angulo AFG. Igitur AFG minor est recto, quare NFB minor recto. Sed OK supra FB facit angulum rectum. Igitur producta concurret cum GN supra BF, et inferius numquam. 
◉ But [angle] AGF = [angle] FGB [by construction], and [angle] FAG > [angle] FBG [because it is subtended by BG, which is longer than AG]. Thus [for the same reason], angle BFG > angle AFG. Accordingly, [angle] AFG is less than a right angle, so [vertical angle] NFB is less than a right angle. But OK forms a right angle on FB. Therefore, when it is extended, it will intersect GN above BF, never below it. 
◉ Facto autem circulo transeunte per tria puncta A, Q, B, transibit supra G, et GQ dividet arcum eius AB per equalia. Sed K dividit cordam AB per equalia. Ergo KO concurret cum GN infra BF et supra punctum G. Igitur prius concurret cum GN infra FB, et iam improbatum est. 
◉ Now, if a circle is produced to pass through the three points A, Q, and B [figure 5.2.44a, p. 265], it will pass above G, and GQ will bisect its arc A[Q]B [since angles AQG and BQG are presumed equal]. But K bisects chord AB [by construction]. Hence, KO will intersect GN below BF and above point G. Therefore, it will intersect GN sooner [than it should, i.e.,] below FB, and it has already been disproven [that it can intersect below FB]. 
◉ Restat ergo ut angulus AQB non sit minor angulo AGD, aut A non reflectitur ad B a puncto Q. Similis erit improbatio, sumpto quolibet puncto arcus EN. 
◉ It therefore follows that angle AQB may not be smaller than angle AGD, or else [the form of point] A is not reflected to B from point Q. The same disproof will hold if any [other] point is taken on arc EN.⁑ 
◉ Sumpto autem puncto in arcu NZ, qui sit C , et fiat reflexio puncti A ad B a puncto C, ut angulus reflexionis supra C sit minor angulo AGD, sicut angulus reflexionis supra T minor eodem, improbatur hoc modo. 
◉ Now, if [some] point C is taken on arc NZ [figure 5.2.44c, p. 267], and if reflection of [the form of] point A to B occurs from point C so that the reflected angle at C is less than angle AGD, just as the reflected angle at T is less than the same angle, [the supposition that the reflected angle at C can be less than angle AGD] is disproved in the following way. 
◉ Ducantur AC, BC, GC. Oportet necessario quod GC dividat KO propter arcum AB, quem dividet ex circulo ABT linea GC per equalia, et similiter linea KO. Sit ergo punctus concursus linee GC cum KO punctus L. Ducta linea TC, cum due linee GC, GT sint equales, erunt duo anguli GCT, GTC equales, et uterque acutus. 
◉ Draw AC, BC, and GC. It follows necessarily that GC intersects KO on arc A[O]B, and line GC will bisect that arc on circle ABT, as will line KO. Accordingly, let point L be where lines GC and KO intersect. When line TC is drawn, then, since the two lines GC and GT are equal [being radii of circle ZHDE], the two angles GCT and GTC will be equal, and both of them will be acute. 
◉ Ducta igitur perpendiculari super GT a puncto T, erit contingens circulo speculi, et producta cadet super terminum dyametri minoris circuli, cum angulus quem efficit cum TG respiciat semicirculum minoris. Et cum TO cadat super KO, et KO producta transeat per centrum minoris circuli, necessario illa perpendicularis cadet super terminum KO producta, et TC est inferior illa perpendiculari, habito respectu ad N. 
◉ Hence, if the line [TX] perpendicular to GT at point T is drawn, it will be tangent to the circle on the mirror, and when it is extended it will fall upon the endpoint of the smaller circle’s diameter [by Euclid, III.31], since the angle it forms with TG is subtended [by an arc equal to] a semicircle on the smaller [circle]. Since TO falls upon KO, and since KO, when extended, passes through the center of the smaller circle [by virtue of its bisecting line AB along the orthogonal], that perpendicular [TX] will necessarily fall upon the endpoint of KO, when it is extended [by Euclid, III.31], and TC lies below that perpendicular when it is taken with respect to N. 
◉ Igitur quecumque linea ducatur ad lineam TC secans dyametrum illius circuli, qui est OK, cadet in punctum linee TC citra illam perpendicularem. Cum igitur GC cadat in C et secet OK, erit C citra perpendicularem et infra arcum illius perpendicularis. 
◉ Therefore, no matter what line is drawn [from G] to line TC so as to intersect diameter OK of that circle, it will fall to a point on line TC but below that perpendicular [TX]. Therefore, since GC falls to C and intersects OK, C will lie below the perpendicular and beneath the arc [on circle AOBT passing] through that perpendicular. 
◉ Facto igitur circulo transeunte per tria puncta A, B, C, transibit quidem per C, et secabit circulum ABT in duobus punctis A, B. Et cum exeat a puncto B, iterum redeat in punctum C, et cum sit citra illum circulum, necessario secabit illum in tertio puncto, quod est impossibile. 
◉ Therefore, if a circle is produced to pass through the three points A, B, and C, it will pass through C, and it will intersect circle ABT at the two points A and B. And when it continues past point B, it will pass on to point C, and although it lies below that circle [i.e., ABT], it will necessarily intersect it at a third point, which is impossible.⁑ 
◉ Restat igitur ut punctus A non reflectatur ad B a duobus punctis arcus interiacentis eorum dyametros, id est arcus EZ, ut uterque angulus reflexionis sit minor angulo AGD, quod est propositum. 
◉ It follows, then, that [the form of point] A may not be reflected to B from two points on the arc extending between the diameters on which they lie, i.e., arc EZ, in such a way that both angles of reflection are less than angle AGD, which is what was proposed. 
◉ [PROPOSITIO 45] Amplius, dico quoniam est reflecti duo puncta a se inequalis longitudinis a centro a duobus punctis arcus ipsa respicientis, id est dyametros in quibus sunt puncta illa interiacentis. 
◉ [PROPOSITION 45] I say, moreover, that two points lying different distances from the center can be reflected from two points on the arc facing them, that is, the arc lying between the diameters in which those points lie. 
◉ Verbi gratia, sumptis duobus dyametris in circulo spere, scilicet BD, GD [FIGURE 5.2.45, p. 603], dividatur angulus eorum per equalia per dyametrum ED, et in BD sumatur punctus M supra punctum in quem cadet perpendicularis ducta a puncto E super BD. Et sumatur ND equalis MD, et fiat circulus transiens per tria puncta D, N, M. Necessario circulus ille transibit extra E, si enim per E, fieret quadrangulum a quatuor punctis D, N, E, M, et duo anguli illius quadranguli sibi oppositi sunt equales duobus rectis, quod quidem non esset, cum linea EM sit supra perpendicularem, et angulus EMD acutus. 
◉ For instance, taking two diameters, i.e., BD and GD [figure 5.2.45, p. 268] in the [great] circle of the sphere, let the angle formed by them be bisected by diameter ED, and take point M on BD beyond the point where the perpendicular dropped from point E to BD will fall. Then take ND equal to MD, and construct a circle passing through the three points D, N, and M. That circle will necessarily pass beyond E, for if [it passed] through E, it would create a quadrilateral at the four points D, N, E, and M, and the two opposite angles of that quadrilateral are equal to two right angles, which would not be so [in this case], since line EM lies beyond the perpendicular, and angle EMD is acute. 
◉ Ei similiter oppositus supra N acutus, quia EN supra perpendicularem. Similis erit improbatio si transeat circulus citra E. Transibit igitur extra, et secabit circulum spere in duobus punctis, sicut T, L. 
◉ Likewise, its counterpart at N is acute, because EN [also lies] beyond the perpendicular [so EMD and END sum up to less than two right angles]. The disproof will be similar if the circle is [assumed] to pass below E. Therefore, it will pass beyond, and it will intersect the [great] circle of the sphere at two points, e.g., T and L. 
◉ Et ducantur linee MT, DT, NT, ML, DL, NL, et ducatur linea MN secans TD in puncto F, lineam ED in puncto P. Palam cum MD sit equalis ND, et PD commune, et angulus equalis angulo, erit triangulus equalis triangulo, et erit angulus FPD rectus. Igitur angulus PFD acutus. 
◉ Draw lines MT, DT, NT, ML, DL, and NL, and draw line MN to intersect [line] TD at point F and line ED at point P. It is evident that, since MD = ND [by construction], since PD is common, and since the [subtended] angle [MDP] = the [subtended] angle [NDP, by construction], triangle [MDP] = triangle [NDP], and angle FPD will be a right angle. Hence, angle PFD is acute. 
◉ Ducatur a puncto F perpendicularis super TD, que sit KF. Palam quoniam aliquis punctus linee NL erit inferior puncto K. Sumpta inferioritate respectu T, sit ille punctus Z, et ducatur linea TZ usque ad circulum cadens in punctum circuli, qui sit C. Arcus NC aut est minor arcu TL, aut non. 
◉ From point F draw KF perpendicular to TD. It is clear that any point on line NL will lie below point K [since FK intersects ML between M and L]. Taking the position of [some such] point [on NL] below [K] with respect to T, let that point be Z, and draw line TZ until it reaches point C on the circle. Arc NC is either shorter than arc TL or not. 
◉ Si non fuerit minor, sumatur ex eo arcus minor, et ad terminum illius arcus ducatur linea a puncto T, et erit idem. 
◉ If it is not shorter, then take an arc on it [i.e., on arc NC] that is shorter, and draw a line from point T to the endpoint of that arc, and it will be the very line [TC that we wanted in the first place]. 
◉ Sit igitur NC minor TL. Palam quoniam angulus TNL erit maior angulo CTN, quia respicit maiorem arcum. Secetur ex eo equalis, et sit INZ, et super punctum T fiat angulus equalis angulo CTN, qui sit OTM. Cum angulus TML sit maior angulo MTO, concurret linea TO cum linea LM. Concurrat in puncto O. 
◉ So let NC < TL. It is obvious that angle TNL > angle CTN, since it is subtended by a longer arc [i.e., TL as opposed to CN]. Cut an [angle] equal [to CTN] from it, let that angle be INZ, and at point T form angle OTM equal to angle CTN. Since angle TML > angle MTO,⁑ line TO will intersect line LM. Let it intersect at point O. 
◉ Cum igitur angulus LMT sit equalis duobus angulis MOT, MTO, et angulus LNT sit equalis angulo LMT, quia super eundem arcum, et angulus INZ sit equalis angulo MTO, erit angulus INT equalis angulo MOT, et ita triangulus MOT similis triangulo INT, et similiter triangulus INZ est similis triangulo TNZ. Et ita proportio NT ad TO sicut NI ad MO, et similiter proportio TN ad TZ sicut IN ad NZ. 
◉ Hence, since angle LMT [exterior to triangle MOT] = [the sum of] the two [interior] angles MOT and MTO [by Euclid, I.32], since angle LNT = angle LMT, because they are subtended by the same arc [TL], and since angle INZ = angle MTO [by construction], angle INT [which = LMT – INZ] = angle MOT [which = LMT – MTO], and so triangle MOT is similar to triangle INT, and likewise triangle INZ is similar to triangle TNZ. So NT:TO = NI:MO, and likewise TN:TZ = IN:NZ. 
◉ Sed TZ maior TO, quod sic patet. Sit R punctus in quo TZ secat KF. Angulus TFR est rectus, quare angulus FTR acutus. Igitur angulus OTF ei equalis est acutus. Et KF perpendicularis super TD, quare producta concurret cum TO, et linea ducta a puncto T ad punctum concursus, cuius linee pars est TO, erit equalis linee TR. Et ita TO minor TZ, quare maior est proportio NT ad TO quam NT ad TZ. 
◉ But TZ > TO, which becomes clear as follows. Let R be the point where TZ intersects KF. Angle TFR is a right angle [by construction], so angle FTR is acute. Hence, angle OTF, which equals it, is acute.⁑ Moreover, KF is perpendicular to TD [by construction], so when it is extended it will intersect TO, and the line [TX] drawn from point T to that point of intersection, a line that includes TO as a segment, will be equal to line TR [because triangles TRF and TXF are equal, by Euclid, I.26]. And thus TO < TZ, so NT:TO > NT:TZ. 
◉ Igitur maior est proportio IN ad MO quam IN ad NZ, quare MO minor NZ. Secetur ergo ex NZ equalis ei que sit NS. 
◉ Therefore [since we have already established that NT:TO = IN:MO and that NT:TZ = IN:NZ], IN:MO > IN:NZ, so MO < NZ. Accordingly, from NZ cut off segment NS equal to MO. 
◉ Quoniam angulus LND cum angulo LMD valet duos rectos, erit angulus LND equalis angulo OMD, et SN, ND equalia OM, MD. Igitur OD equalis est SD. 
◉ Since angle LND + angle LMD = two right angles [by Euclid, III.22], angle LND = angle OMD [which is adjacent to LMD], and SN + ND = OM + MD [since MD = ND, and SN = OM, by construction]. Hence OD = SD [because triangles SND and OMD are equal, by Euclid, I.4]. 
◉ Sed ZD maior SD, quoniam angulus LND cum angulo LMD valet duos rectos. Sed angulus LMD acutus, cum angulus EMD sit acutus. Igitur angulus LND maior est recto. Igitur ZD maior SD, quare ZD maior OD. 
◉ But ZD > SD, because angle LND + angle LMD = two right angles [by Euclid, III.22]. Angle LMD is acute, however, since angle EMD is acute. Therefore, angle LND > a right angle. Hence, ZD > SD [because it intersects LN farther from vertex N than does SD], so ZD > OD [because SD = OD, by previous conclusions]. 
◉ Igitur O refertur ad Z a duobus punctis T, L, et O et Z sunt inequalis longitudinis a centro, et in diversis dyametris. 
◉ Accordingly, [the form of point] O is reflected to Z from the two points T and L [given that OTD = ZTD, by previous conclusions, and ZLD = OLD, since they are subtended by equal arcs], and O and Z lie unequal distances from the center, and they lie on different diameters. 
◉ Et quod non sunt in eodem dyametro palam ex hoc quoniam angulus SDN equalis est angulo ODM. Addito ergo communi angulo SDM, erit angulus NDM equalis angulo SDO. Sed angulus NDM minor duobus rectis, quare angulus ZDO minor duobus rectis. Quare O, Z non sunt in eodem dyametro, sed in diversis. 
◉ That they do not lie on the same diameter is evident from the fact that angle SDN = angle ODM. Thus, when the common angle SDM is added [to each] angle, [angle] NDM = angle SDO. But angle NDM < two right angles, so angle ZDO < two right angles [which means that Z, D, and O cannot lie on a single straight line]. Therefore, O and Z do not lie on the same diameter but on different ones. 
◉ [PROPOSITIO 46] Amplius, sumptis duobus punctis, que sint O, K [FIGURE 5.2.46, p. 604], et inequaliter distantibus a centro, reflectetur quidem unum ad aliud a duobus punctis arcus respicientis semidyametros in quibus sunt, sed non ab alio puncto illius arcus quam ab illis duobus. 
◉ [PROPOSITION 46] Furthermore, given two points O and K [figure 5.2.46, p. 269] that lie different distances from the center [of the mirror], one will be reflected to the other from two points on the arc facing the radii in which those point lie, but they will not reflect from any point on that arc other than those two.⁑ 
◉ Verbi gratia, D sit centrum; K remotior a D quam O a D; GD, OD dyametri; T punctus unus reflexionis. Palam ex superioribus quod duo anguli reflexionis non erunt minores angulo ODA nec equales. Alter ergo erit maior. Sit angulus reflexionis qui est supra T maior, et ducantur linee OT, DT, KT. 
◉ For example, let D be the center [of a great circle on the mirror], let K lie farther from D than O does from D, let GD and OD be diameters, and let T be one point of reflection. It is clear from earlier discussions [in propositions 43 and 44 above] that the two reflected angles will not [both] be smaller than, nor equal to, angle ODA. Hence, one of them will be greater. Let the reflected angle at point T be greater [than angle ODA], and draw lines OT, DT, and KT. 
◉ Et ex angulo illo reflexionis secetur equalis angulo ODA, qui sit OTF, et dividatur angulus FTK per equalia per lineam TE. Et a puncto K ducatur equidistans TF, que quidem concurret cum TE. Concurrat in puncto Z, et ducatur linea OK, et dividatur angulus ODK per equalia per lineam DU secantem lineam OK in puncto C, et sit KD maior OD. Cum igitur sit proportio KD ad DO sicut KC ad CO, erit KC maior CO. Item, linea DT secet lineam OK in puncto N. Dico quoniam C cadit inter N et K non inter N et O, quod sic patebit. 
◉ Then, from that reflected angle [OTK] cut off [angle] OTF equal to angle ODA, and bisect angle FTK with line TE. From point K draw [line KZ] parallel to TF, a line that will intersect TE. Let it intersect at point Z, draw line OK, bisect angle ODK with line DU so that it intersects line OK at point C, and let KD > OD. Therefore, since KD:DO = KC:CO [by Euclid, VI.3], KC > CO. Then, let line DT intersect line OK at point N. I say that C lies between N and K, not N and O, which will be shown as follows. 
◉ Angulus KCD valet duos angulos CDO, COD, et angulus OCD valet duos angulos CKD, CDK. Sed angulus CDO equalis angulo CDK, et angulus KOD maior angulo OKD. Igitur angulus KCD maior angulo OCD, quare angulus KCD maior recto. Et angulus KND acutus, quod sic constabit. 
◉ Angle KCD [exterior to triangle CDO] = the two [interior] angles CDO + COD [by Euclid, I.32], and angle OCD [exterior to triangle CDK] = the two [interior] angles CKD + CDK. But angle CDO = angle CDK [by construction], and [by Euclid, I.19] angle KOD > angle OKD [since KD > OD]. Hence, angle KCD > angle OCD, so angle KCD > a right angle. Also, angle KND is acute, which will be established thus. 
◉ Si fiat circulus per tria puncta O, T, K, transibit infra D, quoniam, cum angulus OTK sit maior angulo ODA, erunt duo anguli OTK, ODK maiores duobus rectis, et linea ND dividet arcum illius circuli, qui est OK, per equalia infra D. 
◉ If a circle is constructed through the three points O, T, and K, it will pass below D, because, since angle OTK > angle ODA [by construction], the two angles OTK + ODK > two right angles, and line ND will bisect arc OK of that circle below D.⁑ 
◉ Si a puncto divisionis ducatur linea ad medium punctum linee OK, que est corda illius arcus, erit linea illa perpendicularis super OK, et cadet inter C et K, cum CK sit maior CO. Et angulus supra N a parte illius perpendicularis et ex parte C erit acutus, et angulus supra C ex parte O est acutus. Si ergo C cadat inter N et O, impossibile erit perpendicularem illam cadere inter N et C, quia secaret DC, et fieret triangulus cuius unus angulus rectus, alius obtusus. 
◉ If a line is drawn from the point of bisection [s] to the midpoint [x] of line OK, which is the chord on that arc, that line will be perpendicular to OK [by Euclid, III.3], and it will fall between C and K, since CK > CO [by previous conclusions]. Moreover, the angle at [point] N beyond that perpendicular on the side of C [i.e., DNC] will be acute, and the angle at C on the side of O [i.e., OCD] is acute [since angle KCD is obtuse, by previous conclusions]. Accordingly, if C were to fall between N and O, it would be impossible for that perpendicular to fall between N and C, because it would intersect DC and would form a triangle with one angle right and the other obtuse.⁑ 
◉ Cadet ergo inter N et K, et erit angulus N ex parte perpendicularis acutus; igitur ex parte C obtusus, et ita erit triangulus cuius duo anguli obtusi. 
◉ Hence, it [i.e., the perpendicular] will fall between N and K, and the angle at N on the side of the perpendicular will be acute, so this [same angle] will be obtuse on the side of C [if C lies between N and O], and so there will be a triangle with two obtuse angles.⁑ 
◉ Palam quoniam angulus KTD est medietas anguli KTO, sed KTE medietas anguli KTF. Restat ETD medietas anguli FTO, sed FTO equalis est angulo ODA. Igitur ETD medietas anguli ODA. 
◉ It is evident that angle KTD is half of angle KTO [by construction], but [angle] KTE is half of angle KTF [by construction]. It follows that [angle] ETD is half of angle FTO [since ETK = FTK, by construction], but [angle] FTO = angle ODA [by construction]. Thus, [angle] ETD is half of angle ODA. 
◉ Sed angulus ODA cum angulo ODF valet duos rectos, et tres anguli trianguli ETD duos rectos. Ablato EDT communi, restat angulus TED equalis medietati anguli ODA et angulo ODN. Sed angulus ODC cum medietate anguli ODA est rectus. Igitur angulus TED est acutus, quare ei contrapositus est acutus. 
◉ But angle ODA + [adjacent] angle ODF = two right angles, and the three angles of triangle ETD sum up to two right angles. With common [angle] EDT subtracted, [angle] TED remains equal to half of angle ODA + angle ODN. But angle ODC + half of angle ODA is a right angle [since 2ODC (which = ODK) + ODA = two right angles]. Therefore, angle TED is acute [since TED = ODC – NDC + half of ODA], so its vertical angle [KEZ] is acute. 
◉ Igitur, si a puncto K ducatur perpendicularis ad TZ, cadet inter E et Z. Si enim supra E ceciderit, cum angulus TEK sit obtusus, accidet triangulum habere duos angulos rectum et obtusum. Sit ergo perpendicularis KQ. Dico quoniam KT se habet ad TF sicut KD ad DO. [2.449] Probatio: TO aut est equidistans KD, aut concurrit cum ea. Sit equidistans [FIGURES 5.2.46a, 5.2.46b, p. 605]. Erit angulus ODA equalis angulo TOD, et ita TOD equalis angulo OTF. OD, TF aut sunt equidistantes, aut concurrent. 
◉ Thus, if a perpendicular is dropped from point K to TZ, it will fall between E and Z. For if it were to fall above E, then, since angle TEK is obtuse, it would follow that the triangle [formed by KD and the perpendicular intersecting TE above E] would have two [of its three] angles [consisting of] a right angle [formed by the line from K that intersects TE above E] and an obtuse angle [KET]. Let KQ be the perpendicular, then. I say that KT:TF = KD:DO. 
◉ Si equidistantes [FIGURE 5.2.46a, p. 605], cum cadant inter equidistantes, erunt equales. Si vero concurrunt [FIGURE 5.2.46b, p. 605], faciunt triangulum cuius latera equalia, quia respiciunt equales angulos, et FD secat illa latera equidistans basi. Erit ergo proportio unius laterum ad DO sicut alterius ad FT, et ita TF equalis DO. 
◉ If they are parallel [figure 5.2.46a, p. 270], then, since they fall between parallels, they will be equal. If, however, they intersect [figure 5.2.46b, p. 270], then they form a triangle [two of] whose sides [OP and TP] are equal, because they subtend equal angles [since angle DOT = alternate angle ODA, and angle OTF = angle ODA, by construction], and because FD intersects those sides parallel to the base [TO]. Therefore, the ratio of one side to DO will be the same as the ratio of the other side to TF, and so TF = DO. 
◉ Et hoc dico si concurrant sub KD. Et si concurrant sub TO eadem erit probatio, quia fiet triangulus cuius unum latus TO et alia duo latera equalia, et erit proportio unius ad DO sicut alterius ad TF. Item, angulus TDK equalis angulo DTO, quia DT inter equidistantes. Igitur est equalis angulo DTK, quare DK, TK sunt equalia. Igitur proportio TK ad TF sicut KD ad DO. 
◉ {I claim that this is so if they [i.e., DO and TF] intersect below KD. And if they intersect below TO, the same proof will apply, because they will form a triangle one of whose sides is TO and the other two of which are equal, and the ratio of one [of those equal sides] to DO will be the same as the ratio of the other to TF.}⁑ Furthermore, angle TDK = [alternate] angle DTO, because DT lies between parallels. Thus, it [i.e., angle TDK] = angle DTK [which = angle of reflection DTO], so DK and TK are equal. Hence, TK:TF = KD:DO. 
◉ Si vero TO concurrit cum KD, concurrat ex parte A in puncto P [FIGURE 5.2.46c, p. 605]. Scimus quoniam proportio KT ad TF compacta est ex proportione KT ad TP et TP ad TF. Sed proportio KT ad TP sicut KD ad DP, quoniam DT dividit angulum KTO per equalia. Et proportio TP ad TF sicut DP ad DO, quoniam angulus ODP equalis angulo PTF, et angulus supra P communis. Erit partialis triangulus similis totali. Igitur proportio KT ad TF constat ex proportione KD ad DP et proportione DP ad DO. Sed proportio KD ad DO constat ex eisdem, quare proportio KT ad TF sicut KD ad DO. 
◉ If, on the other hand, TO intersects KD, let it intersect at point P on the side of A [figure 5.2.46c, p. 270]. We know that KT:TF is compounded of KT:TP and TP:TF [i.e., KT:TF = (KT:TP):(TP:TF)]. But KT:TP = KD:DP [by Euclid, VI.3], because DT bisects angle KTO. Moreover [given that triangles TPF and DPO are similar], TP:TF = DP:DO, because angle ODP [i.e., ODA] = angle PTF [by construction], and the angle at P is common. Part [ODP] of the [larger] triangle [PTF] is [therefore] similar to the whole. Hence, KT:TF is compounded of KD:DP [which = KT:TP] and DP:DO [which = TP:TF]. But KD:DO is compounded of the same [ratios, i.e., KD:DP and DP:DO], so KT:TF = KD:DO. 
◉ Si vero TO concurrat cum KD ex parte G [FIGURE 5.2.46d, p. 605], sit concursus L, et a puncto D ducatur equidistans linee KT, que sit DR, concurrens cum TO in puncto R. Igitur angulus KTD equalis est angulo TDR, sed idem est equalis angulo DTO, quare DR est equalis TR. Sed quoniam triangulus LTK similis triangulo LRD, erit proportio DR ad RL sicut KT ad TL, et ita RT ad RL sicut KT ad TL. Sed RT ad RL sicut DK ad DL. Igitur KT ad TL sicut KD ad DL. 
◉ If, however, TO intersects KD on the side of G [figure 5.2.46d, p. 270], let L be the [point of] intersection, and from point D draw [line] DR parallel to line KT so as to intersect TO at point R. Accordingly, angle KTD = [alternate] angle TDR, but it is [also] equal to angle DTO [by supposition], so DR = TR. However, since triangle LTK is similar to triangle LRD [because TK and RD are parallel], DR:RL = KT:TL, and so RT [which = DR]:RL = KT:TL. But RT:RL = DK:DL. Thus, KT:TL = KD:DL. 
◉ Sed quoniam angulus FTO equalis est angulo ODA, erit angulus ODL equalis angulo FTL, et angulus supra L communis. Erit triangulus ODL similis triangulo FTL. Igitur TL ad TF sicut DL ad DO, et ita KT ad TL sicut KD ad DL. Et TL ad TF sicut DL ad DO, quare KT ad TF sicut KD ad DO, quod est propositum. 
◉ Yet, since angle FTO = angle ODA [by construction], angle ODL [adjacent to ODA] = angle FTL [adjacent to FTO], and the angle at L is common. [Thus], triangle ODL will be similar to triangle FTL. Hence, TL:TF = DL:DO, and so [by previous conclusions] KT:TL = KD:DL. Moreover [by previous conclusions], TL:TF = DL:DO, so KT:TF = KD:DO, which is what was proposed.⁑ 
◉ Sed quoniam KZ [FIGURE 5.2.46, p. 604] equidistans TF, erit angulus KZE equalis angulo ETF, et ita triangulus KZE similis triangulo ETF, quare proportio KE ad EF sicut KZ ad TF. Sed KE ad EF sicut KT ad TF, propter angulum supra T divisum per equalia. Igitur KZ equalis KT. 
◉ However, since KZ [in figure 5.2.46, p. 269] is parallel to TF [by construction], angle KZE = [alternate] angle ETF [and vertical angle ZEK = vertical angle TEF], so triangle KZE is similar to triangle ETF, and so KE:EF = KZ:TF. But [by Euclid, VI.3] KE:EF = KT:TF, because the angle at T is bisected. Thus, KZ = KT. 
◉ Verum quoniam KQ est perpendicularis super EZ, erunt omnes eius anguli recti. Sed angulus ETD est acutus, quoniam est medietas anguli. Igitur KQ concurret cum TD. Sit concursus H, et ducatur linea EH, et a puncto E ducatur equidistans KH producta usque ad DH, que sit EC. 
◉ Since KQ is perpendicular to EZ, though, all of its angles [of intersection with TZ] will be right angles. But angle ETD is acute, because it is half of angle [FTO, by earlier conclusions]. Therefore, KQ will intersect TD. Let H be the intersection, draw line EH, and from point E draw EC’ parallel to KH, and extend it to DH [which it intersects at point C’]. 
◉ Et mutetur figura propter intricationem linearum [FIGURE 5.2.46e, p. 606], et fiat circulus transiens per tria puncta C, T, E. Et producatur KD usque in circulum cadens in punctum M, et ducatur MT. Erit angulus TME equalis angulo TCE, quia cadunt in eundem arcum, et angulus TCE equalis angulo CHK. Erit TME equalis angulo CHK. 
◉ Adjust the figure according to the interrelationship of lines [figure 5.2.46e, p. 271], and construct a circle that passes through the three points C’, T, and E. Extend KD until it reaches the circle at point M, and draw MT. Angle TME = angle TC’E, since they are subtended by the same arc, and angle TC’E = angle C’HK [since EC’ and KH are parallel, by construction]. [Therefore, angle] TME = angle C’HK. 
◉ Secetur ab angulo TME angulus equalis angulo DHE, qui sit FMD, et punctus in quo FM secat TC sit I. Palam quoniam triangulus IMD similis est triangulo EDH, quare proportio HD ad DM sicut EH ad IM. 
◉ Cut from angle TME angle F’MD equal to angle DHE, and let I be the point where F’M intersects TC’. It is evident that triangle IMD is similar to triangle EHD [because all their corresponding angles are equal], so HD:DM = EH:IM. 
◉ Et similiter, triangulus TMD similis triangulo KHD, et proportio KD ad DT sicut HD ad DM, et ita KD ad DT sicut EH ad IM. 
◉ Likewise, triangle TMD is similar to triangle KHD [because angle TMD (i.e., TME) = angle TC’E (which = angle THK), by construction, and angle HDK = vertical angle TDM], and [therefore, in similar triangles TMD and KHD, the corresponding sides are proportional, so] KD:DT = HD:DM, and so [given that HD:DM = EH:IM, by previous conclusions] KD:DT = EH:IM. 
◉ Sed proportio KD ad DT nota, quoniam semper una et eadem permanet, cuicumque punctus reflexionis sit T in arcu EG, quia semper linea TD una, et KD similiter. Linea etiam EH una in quacumque reflexione permanet, et non mutatur eius quantitas, quare linea IM semper erit una, quare punctus F notus et determinatus. 
◉ But KD:DT is known, since it remains one and the same throughout, no matter where point of reflection T might lie on arc EG, because TD remains unchanged, and so does KD. Line EH also remains one [and the same] no matter the reflection, so it does not change its length, and so line IM will always be one [and the same], and so point F’ is known and determinate.⁑ 
◉ Si ergo a tribus punctis arcus BG fieri posset reflexio, esset ducere a puncto F ad circulum TCE tres lineas equales, quia esset proportio KD ad DT sicut EH ad quamlibet illarum. Et patet ex superioribus quod non nisi due equales duci possunt, quare a duobus tantum punctis fiet reflexio, quod est propositum. 
◉ Therefore, if reflection could occur from three points on arc BG, three lines equivalent [to FM] could be extended from point F’ to circle TC’E, because KD:DT would be as EH is to any one [segment IM] of them. But it is clear from earlier discussion [in proposition 20, lemma 2 above] that only two equal lines can be [so] drawn, so reflection will occur from only two points, which is what was proposed.⁑ 
◉ [PROPOSITIO 47] Amplius, datis duobus punctis K, O [FIGURE 5.2.47, p. 607] in diversis dyametris inequaliter distantibus a centro, est invenire punctum reflexionis. 
◉ [PROPOSITION 47] Furthermore, given two points K and O [figure 5.2.47, p. 275] lying on different diameters and at different distances from the center, to find [either] point of reflection. 
◉ Verbi gratia, sumatur linea ZT, et dividatur in puncto E ut sit proportio ZE ad ET sicut KD ad DO. Quoniam KD maior DO, erit ZE maior ET. Dividatur ZT per equalia in puncto Q, et a puncto Q ducatur perpendicularis super ZT, et fiat angulus ETD equalis medietati anguli ODA. Erit quidem acutus. Igitur TD concurret cum perpendiculari. 
◉ For example, take line ZT’, and cut it at point E so that ZE:ET’ = KD:DO. Since KD > DO [by construction from the previous proposition], ZE > ET’. Bisect ZT’ at point Q, from point Q draw a line [K’QH] perpendicular to ZT’, and form angle ET’D’ equal to half of angle ODA. It will be acute. Accordingly, T’D’ will intersect the perpendicular [KQ]. 
◉ Sit concursus in puncto H, et ducatur linea DEK ut sit proportio KD ad DT sicut KD ad semidyametrum spere. Et angulo quem habemus KDT fiat in speculo angulus equalis KDT. Dico quoniam T est punctus reflexionis, et si predictam probationem replicaveris, manifeste videbis. 
◉ Let the intersection be at point H, and [by proposition 24, lemma 6] draw line D’EK’ so that K’D’:D’T’ = KD:[DT], the radius of the sphere. Then, form angle KDT in the mirror equal to angle K’D’T’ that we have [in the construction based on T’Z]. I say that T is a point of reflection, and if you repeat the previous proof, you will see this clearly.⁑ 
◉ [PROPOSITIO 48] Amplius, sumptis duobus punctis in diversis dyametris, que puncta inequalis sint longitudinis a centro, si fuerint extra circulum, et reflectantur ab aliquo puncto arcus oppositi dyametris, non reflectentur ab alio eiusdem arcus. 
◉ [PROPOSITION 48] Moreover, if two points are taken on different diameters, if they lie at different distances from the center, and if they lie outside the circle and are reflected from some point on the [concave] arc facing the diameters, they will not be reflected from any other [point] on the same arc. 
◉ Verbi gratia, sint A, B [FIGURE 5.2.48, p. 607] puncta in diversis dyametris extra circulum, G centrum, T punctus reflexionis, et ducantur BT, AT, GT. BT secabit arcum circuli. Sit punctus sectionis Q. AT secabit similiter arcum circuli. Sit punctus sectionis M. 
◉ For example, let A and B [figure 5.2.48, p. 276] be points lying outside the circle on different diameters, G the center [of the great circle on the mirror], and T the point of reflection, and draw BT, AT, and GT. BT will intersect the [convex] arc on the circle. Let Q be the point of intersection. So too, AT will intersect the [convex] arc on the circle. Let M be the point of intersection. 
◉ Quoniam angulus BTG equalis est angulo ATG, cadunt in arcus circuli equales, quod patebit, producto dyametro TG. Erit ergo arcus QT equalis arcui MT. Si igitur B refertur ad A ab alio puncto, sit illud H, et ducantur linee BH, AH, GH. Secet BH circulum in puncto L, AH in puncto N. 
◉ Since angle BTG = angle ATG [by construction], they are subtended by equal arcs on the circle, which will be evident if diameter TG is drawn. Accordingly, arc QT = arc MT [and so, therefore, do chords QT and MT]. Thus, if [the form of point] B is reflected to [point] A from a point other [than T], let it be H, and draw lines BH, AH, and GH. Let BH intersect the circle at point L, AH at point N. 
◉ Secundum supradictam rationem, erit HL equalis NH. Sed iam habemus quod QT equalis TM, quod est impossibile. Restat ut B non reflectatur ad A a puncto H vel ab alio puncto arcus oppositi dyametri preter quam T. 
◉ According to the previous reasoning, HL = NH [because they subtend supposedly equal arcs]. But we have just [established] that QT = TM, which is impossible [if HL = NH]. It follows that [the form of point] B may not reflect to [point] A from point H or from any point other than T on the arc facing [either] diameter. 
◉ Similiter, si fuerit alterum punctorum in circulo, alterum extra, ab uno tantum puncto arcus poterit reflecti ad aliud. 
◉ Likewise, if one of the points lies on the circle, while the other lies outside, then the one can be reflected to the other from only one point on the arc. 
◉ Amplius, si linea ducta ab uno duorum punctorum ad aliud contingat circulum, aut tota sit extra, sumpto quocumque puncto in arcu opposito dyametris, altera linearum a punctis duobus ad illud punctum ductarum tota erit extra circulum. Et sic neuter punctorum ad alium reflectetur ab aliquo puncto illius arcus, et ab uno solo puncto speculi arcus oppositi reflectetur, et ita ab uno solo puncto speculi. 
◉ Moreover, if the line extending from one of the two points [A or A’ in figure 5.2.48a, p. 276] to the other [B] is tangent to the circle or lies entirely outside it, then, if some point [T] is taken on the [convex] arc facing the diameters, one of the lines [i.e., either AT or A’T] extended from [each of] the two points to that point will lie entirely outside the circle [because it will strike its outer rather than its inner surface]. And so neither of the points [A and B, or A’ and B] will be reflected to the other from any point on that arc [CE], and [it will be reflected] from only one point on the opposite arc [KD] of the mirror, and so [it will be reflected] from only one point on the [entire] mirror. 
◉ [PROPOSITIO 49] Si vero linea ducta ab uno puncto ad alium secet circulum, fiat circulus super centrum speculi et illa duo puncta. Circulus ille aut totus erit intra circulum, aut continget ipsum, aut secabit. 
◉ [PROPOSITION 49] On the other hand, if the line extending from one point to the other [i.e., from objectpoint to center of sight] cuts the [great] circle [on the mirror], construct the circle [passing] through the center of the mirror and those two points. That [second] circle will lie entirely within the circle [of the mirror], or it will touch it [at one point], or it will intersect it. 
◉ Sit totus intra, et ducantur due linee a duobus punctis ad aliquod punctum arcus oppositi. Angulus quem facient erit minor angulo quem unus dyameter facit cum alio ex alia parte centri, et quilibet angulus sic factus super arcum oppositum minor erit illo angulo. 
◉ Let it lie entirely within [figure 5.2.49, p. 277], and draw two lines [AT and BT] from the two points to some point [T] on the facing arc [of the mirror]. The angle they will form [ATB] will be smaller than the angle one diameter forms with the other on the adjacent side of the [mirror’s] center [i.e., angle DGB]; and no matter what angle is formed in this way on the facing arc, it will be smaller than the latter angle. 
◉ Quoniam angulus factus in interiori circulo per lineas a punctis ad arcum eius interiacentem ductas erit equalis illi angulo, quoniam cum angulo dyametrorum supra centrum valet duos rectos. Sed angulus arcus minoris circuli maior angulo arcus speculi. 
◉ For the angle [ACB] formed within the inner circle by the lines drawn from the points to the arc on it lying between [the two points] will be equal to that latter angle [DGB], because, combined with the angle [AGB] formed by the diameters above the center, it sums up to two right angles [by Euclid, III.22]. But the angle [ACB] within arc [ACB] in the smaller circle is greater than the angle [ATB] within arc [CTH of the circle] on the mirror. 
◉ Igitur in arcu circuli non fiet reflexio nisi ab uno puncto, cum iam dictum sit quod non est possibile reflexionem duobus punctis fieri ut sit uterque angulus minor angulo dyametrorum ex alia parte centri. 
◉ Therefore, in the arc of the circle [on the mirror], reflection will occur from only one point, since it has already been claimed [in proposition 44 above] that it is not possible for reflection to occur from two points such that both [reflected] angles are smaller than the angle formed by the diameters on the adjacent side of the center. 
◉ Si vero circulus ille contingat circulum speculi, angulus factus a lineis ab illis punctis ad punctum contactus ductis erit equalis angulo dyametrorum ex alia parte centri, quare ab illo punto contactus non fiet reflexio. Et angulus factus super quodcumque punctum aliud arcus maioris circuli erit minor illo, quare a duobus punctis arcus non fiet reflexio, secundum predicta. 
◉ If, however, that [second] circle touches the circle of the mirror at one point [T in figure 5.2.49a, p. 277], the angle formed by the lines drawn from those points [A and B] to the point of contact [T] will be equal to the angle [DGB] formed by the diameters on the adjacent side of the center [because AGB + DGB = two right angles, by Euclid, III.22, and so do AGB + ATB], so no reflection will occur from that point of contact [according to proposition 43 above]. Moreover, the angle formed at any other point [e.g., T’] on the arc [CH] of the larger circle will be smaller than that one, so, according to previous claims [in proposition 44 above], reflection will not occur from two points on that arc. 
◉ Si vero circulus interior secet circulum speculi, duo puncta aut erunt extra circulum; aut intra; aut unus intra, alius extra; aut unus in circulo, alius extra vel intra. 
◉ If, on the other hand, the inner circle cuts the circle of the mirror, the two points [A and B] will lie outside the circle [of the mirror], or [they will both lie] inside it, or one [will lie] inside and the other outside, or one [will lie] on the circle and the other outside or inside it. 
◉ Si fuerint extra, vel unus in circulo alius extra, circulus secans non secabit arcum circuli speculi interiacentem dyametris, et ita quilibet angulus factus super arcum illum erit maior angulo dyametrorum ex alia parte centri. Et iam probatum est in precedenti figura quod hec puncta ab uno solo puncto arcus interiacentis poterunt reflecti. 
◉ If they [both] lie outside [as represented by A and B in figure 5.2.49b, p. 277], or if one lies on the circle and the other outside [as represented by A and B’ in the same figure], then the [second] intersecting circle will not cut the arc of the mirror’s circle between the diameters, and so, no matter what angle is formed on that arc [e.g., ATB or ATB’], it will be larger than the angle [DGB] formed by the diameters on the adjacent side of the center. And it has already been demonstrated in the preceding figure [i.e., proposition 48] that [the forms of] these points can be reflected from only one point on the [concave] arc [KD] lying between [the diameters].⁑ 
◉ Si vero duo puncta fuerint intra, secabit circulus interior arcum interiacentem in duobus punctis, et restabunt ex eo duo arcus ex diversis partibus. 
◉ But if the two points lie inside [the mirror’s circle, as represented in figure 5.2.49c, p. 277], the inner circle will cut the arc lying between [the diameters] at two points [E and F], and there will be two arcs [CE and FH] left over on opposite sides [of arc CH facing the diameters]. 
◉ Si unus punctorum fuerit intra circulum, alius in circulo vel extra, secabit circulus arcum interiacentem in unico puncto, et restabit unus arcus tantum. 
◉ If one of the points lies inside the circle and the other on the circle or outside it [as represented in figure 5.2.49d, p. 278], the [second] circle will cut the arc [CB’] lying between [the diameters] at a single point [E], and only one arc[segment, CE] will be left over. 
◉ Si secet in duobus punctis, omnes anguli facti super arcum interiacentem duo puncta sectionis erit maior angulo dyametrorum ex alia parte centri, et ab hoc arcu poterit fieri reflexio forsitan ab uno puncto tantum, forsitan a duobus. 
◉ If it intersects [the arc between the diameters] at two points, all the angles [e.g., ATB in figure 5.2.49c, p. 277] formed upon the arc lying between the two points of intersection [E and F] will be greater than the angle [DGB] formed by the diameters on the adjacent side of the center, and from this arc [EF] reflection may occur from only one point, or it may occur from two [by proposition 46 above]. 
◉ Et a duobus arcubus qui restant ex arcu totali, et ex diversis partibus, omnes anguli erunt minores angulo dyametrorum, et tantum ab uno eorum puncto fiet reflexio. 
◉ And from the two arcs [CE and FH] that are left over from the entire arc on opposite sides of [points E and F of intersection], all the angles [e.g., AT’B] will be smaller than the angle [DGB] formed by the diameters [on the adjacent side of the center], and reflection will occur from only one point on them [by proposition 44 above]. 
◉ Et in hoc situ poterit fieri reflexio a duobus punctis arcus interiacentis dyametros, aut a tribus. 
◉ And [so] in this case [when two arcsegments are left over on opposite sides of intersectionpoints E and F], reflection can occur from two points on the arc lying between the diameters, or from three points [in the whole arc CH, i.e., two from EF and one from CE or FH]. 
◉ Et palam quod ab uno tantum puncto arcus oppositi fiet reflexio, et ita in hoc situ aliquando a tribus, aliquando a quatuor. 
◉ Moreover, it is clear [from proposition 38 above] that reflection will occur from only one point on the opposite arc [DK], so in this case it may occur from three [points], or it may occur from four.⁑ 
◉ Si vero secetur arcus interiacens dyametros in uno tantum puncto a maiori circulo, omnes anguli facti in parte illius arcus inclusa minori circulo erunt maiores angulo dyametrorum, et poterit fieri reflexio a duobus punctis illius partis, vel ab uno. 
◉ But if it [i.e., the second circle] cuts the arc lying between the diameters at only one point on the larger circle, all the angles formed on the segment of that arc [EB’ in figure 5.2.49d, p. 278] included within the smaller circle will be greater than the angle [DGB’] formed by the diameters [on the adjacent side of the center], and [so] reflection can occur from two points, or from one point, on that part [CB’ of the arc facing the diameters]. 
◉ Omnes anguli alterius partis arcus interiacentis erunt minores angulo dyametrorum, et ab uno puncto tantum illius partis fiet reflexio, et ita, cum ab uno puncto arcus oppositi semper fiat reflexio in hoc situ, aliquando a tribus, aliquando a quatuor, non a pluribus poterit esse reflexio. 
◉ All the angles in the other side [CE] of the arc lying between [C and B’] will be smaller than the angle formed by the diameters [on the adjacent side of the center], and reflection will occur from only one point on that segment, and so, given that reflection always occurs in this case from one point on the opposite arc [DK of the mirror], reflection may occur from three [points], or it may occur from four, but never from more. 
◉ Palam ergo quod puncta inequalis longitudinis a centro, aliquando ab uno puncto tantum, aliquando a duobus, aliquando a tribus, aliquando a quatuor, numquam a pluribus, reflectuntur. Cum autem puncta eiusdem longitudinis fuerint, poterit fieri reflexio aut ab uno tantum puncto, aut a duobus, aut a quatuor, numquam a tribus. 
◉ It is therefore clear that points lying different distances from the center may at times be reflected from only one point, at times from two, at times from three, or at times from four, but never from more [than four]. Moreover, if the [two] points lie the same distance [from the center], reflection can occur from one point only, or from two, or from four, [but] never from three [alone].⁑ 
◉ Ubi ab uno puncto fit reflexio, una apparet ymago; ubi due, due; ubi tres, tres; ubi quatuor, quatuor. Si vero punctus visus et centrum visus fuerint in eodem dyametro, fiet reflexio a circulo toto, et locus ymaginis erit centrum visus. Verum, si centrum visus fuerit in centro speculi, nichil videt. Si vero punctus visus fuerit in centro speculi, non videbitur, quoniam forma eius accedet ad speculum super perpendicularem, nec reflecti poterit nisi super perpendicularem. 
◉ When reflection occurs from one point, one image appears; when [it occurs] from two, two [images appear]; when [it occurs] from three, three [images appear]; when [it occurs] from four, four [images appear]. But if the visible point and the center of sight lie on the same diameter, reflection will occur from an entire circle [within the sphere of the mirror], and the imagelocation will be [at] the center of sight [by proposition 36 above]. If the center of sight lies at the center of the mirror, though, it sees nothing [other than itself]. On the other hand, if the visible point lies at the center of the mirror, it[s image] will not be seen, because its form will reach the mirror along the normal and can be reflected only along the normal. 
◉ Cum autem centrum visus et punctus visus fuerint in diversis lineis extra centrum, linee ille ad centrum producte secabunt in diversis partibus ex circulo spere duos arcus. Ab uno puncto unius tantum fiet reflexio, ab alio forsitan a tribus. Quod si centrum spere fuerit ex una parte, centrum visus et punctus visus ex una, arcus quem secant dyametri propter oppositionem capitis abscondetur, unde tunc a tribus tantum punctis fiet reflexio. Et si dirigatur in hoc situ visus ad arcum unius reflexionis tantum, abscondetur alius trium, et unica apparebit ymago. [2.489] Item, si integrum fuerit speculum, non erit ibi perceptio. Oportet igitur ut in eo sit abscisio, et accidet non numquam arcum interiacentem dyametros abscisum esse, et tunc nichil in eo videri, quare raro eveniet quatuor ymagines in hoc speculo comprehendi. Unde si quis hanc pluralitatem ymaginum voluerit videre, disponat visum intra speculum circa ipsum ut modicam partem eius abscondat mole capitis, et totam speculi superficiem visu discurrat. 
◉ Yet when the center of sight and the visible point lie outside the center on different lines, those lines, when extended to the center, will cut two arcs on different sides of the circle within the sphere. Reflection will occur from only one point on one [of those arcs], but [it may occur] from three [points] on the other. But if the center of the sphere lies on one side, while the center of sight and the visible point lie on the other, the arc that the diameters cut [on one side or the other in the circle] will be blocked by the [viewer’s] head, so reflection will then occur from only three points [at most]. And if in that case the eye is directed to the arc where only a single reflection occurs, the other arc [from which] three [reflections occur] will be blocked, and only one image will appear. 
◉ Cum autem aliquid in hoc speculo percipietur duplici visu, si linea reflexionis fuerit equidistans perpendiculari, erit locus ymaginis punctus reflexionis, et cum distent a se puncta reflexionis respectu duorum visuum, apparebunt duobus visibus due ymagines eiusdem puncti. Si vero linea reflexionis non sit equidistans perpendiculari, et punctus visus tantum distet ab uno visu quantum ab alio, vel modica sit differentia, erit locus ymaginis respectu utriusque visus idem, aut diversus, sed modicum distans. Unde aut una apparebit ymago, aut fere una, sicut probatum est in speculis spericis exterioribus. 
◉ When something is perceived in this [sort of] mirror with both eyes, if the line of reflection is parallel to the normal, the imagelocation will lie at the point of reflection [by proposition 32 above], and since the points of reflection are separated from one another with respect to the two eyes, two images of the same point will appear to the two eyes [and will be melded at the point of reflection]. On the other hand, if the line of reflection is not parallel to the normal, and if the visible point lies the same distance from one eye as from the other, or if the difference [in distance] is slight, the imagelocation will be the same for both eyes, or it will be different [for each eye] but only slightly divergent. Therefore, either a single image, or virtually a single image, will appear, as was demonstrated in the case of convex spherical mirrors [in paragraph 2.221 above]. 
◉ In speculis columpnaribus concavis, aliquando linea communis est linea recta. Cum superficies reflexionis transit per axem, aliquando linea communis est circulus—cum superficies illa est equidistans basibus—aliquando linea communis est sectio columpnaris. Quando fuerit linea recta, erit locus ymaginis et modus reflexionis sicut in speculis planis. Quando fuerit circulus, erit idem modus qui in concavis spericis. Cum vero fuerit columpnaris sectio, aut erit locus ymaginis ultra speculum, aut citra visum, aut in centro visus, aut inter speculum et visum, aut in ipso speculo, quod sic patebit. 
◉ In the case of concave cylindrical mirrors, the common section [of the plane of reflection and the mirror] is sometimes a straight line. When the plane of reflection intersects the axis, the common section is sometimes a circle—[i.e.,] when that plane is parallel to the bases [of the cylinder]—[and] the common section is sometimes a cylindric section [i.e., an ellipse]. When [the common section] is a straight line, imagelocation and the analysis of reflection will be the same as in plane mirrors. When it is a circle, the analyis will be the same as in concave spherical [mirrors]. However, when the [common] section is cylindric [i.e., elliptical], the imagelocation will lie behind the mirror, or beyond the center of sight, or at the center of sight, or between the mirror and the center of sight, or on the mirror itself, which will be demonstrated in the following way. 
◉ [PROPOSITIO 50] Sit ABG [FIGURE 5.2.50, p. 608] sectio. Ducatur perpendicularis in hac sectione, que sit DG, quam secundum predicta patet esse dyametrum circuli, et unicam posse esse, cum ab alio puncto sectionis non possit duci perpendicularis super superficiem contingentem. Sumatur aliud punctum, et sit B, et ducatur ab eo in sectione linea perpendicularis super lineam contingentem sectionem in puncto B, que quidem linea, secundum predicta, necessario concurret cum perpendiculari. Concurrat in puncto D, et sumatur B circa punctum G ut angulus BDG sit acutus. 
◉ [PROPOSITION 50] Let ABG [figure 5.2.50, p. 280] be the [elliptical common] section. Draw normal DG within this section, and, according to previous discussion [in book 4], it is clear that this normal is a diameter of the circle [parallel to the cylinder’s base and coincident with the section], and it must be unique, because from no other point on the section can a normal be drawn to the plane tangent [to both the circle and the section].140 Select another point [on the section], let it be B, and from it draw a line within the section that is normal to the line tangent to the section at point B, and, as claimed earlier [in proposition 26], this line will necessarily intersect normal [GD]. Let it intersect at point D, and let [point] B be chosen near enough to point G that angle BDG is acute. 
◉ Deinde a puncto G ducatur in sectione linea equidistans BD, que sit GH, que quidem cadat intra columpnarem sectionem, quia erit angulus HGD acutus, cum sit equalis GDB. Et a puncto G inter D et H ducatur linea, que necessario concurret cum BD. Concurrat in puncto N, et inter N et G sumatur punctus quicumque, qui sit O. Ultra punctum N sumatur punctum T. Item, a puncto G ducatur supra GH alia linea GZ tamen intra sectionem, que necessario concurret cum BD ex alia parte. Sit concursus E. Ducatur GQ linea ut angulus QGD sit equalis angulo ZGD, et fiat angulus LGD equalis angulo HGD, et angulus MGD equalis angulo NGD. 
◉ Then, from point G, draw line GH parallel to BD within the section, and it should lie within the cylindric section, because angle HGD will be acute, since it is equal to [alternate angle] GDB. From point G draw a line between D and H, and it will necessarily intersect BD. Let it intersect at point N, and between N and G select some point O. Beyond point N [on line GN] select point T. Furthermore, from point G draw another line GZ above GH, [but] still within the section, and it will necessarily intersect BD on the other side [of G]. Let E be the [point of] intersection. Draw line GQ so that angle QGD = angle ZGD, and form angle LGD equal to angle HGD and angle MGD equal to angle NGD. 
◉ Palam quod, si fuerit visus in puncto Z, reflectetur punctus Q ad ipsum a puncto G, et punctus ymaginis E. Et si fuerit visus in puncto H, reflectetur ad ipsum punctus L a puncto G, et erit locus ymaginis G. Si vero fuerit visus in puncto O, reflectetur ad ipsum punctus M, et locus ymaginis N. Si autem fuerit in N, erit locus ymaginis puncti M in centro visus, id est in N. Si autem fuerit in T, erit locus ymaginis inter visum et speculum, quia in N, et ita propositum. 
◉ It is evident that, if the center of sight lies at point Z, [the form of] point Q will be reflected to it from point G [since QGD = ZGD by construction], and point E [behind the mirror on normal QD will be the location] of its image. If the center of sight lies at point H, [the form of] point L will be reflected to it from point G [since LGD = HGD by construction], and its imagelocation will be [point] G [on the mirror’s surface, because normal LD is parallel to line of reflection GH]. If the center of sight is at point O, [the form of] point M will be reflected to it [from point G, since MGD = NGD by construction], and its imagelocation will be [point] N [behind the eye on normal MD]. If [the center of sight] lies at N, the imagelocation [for the form] of point M will be at the center of sight [itself], i.e., at N [where normal MD intersects line of reflection GN]. And if [the center of sight] is at T, the imagelocation [for point M] will lie between the eye and the mirror, because it lies at N, and so [we have demonstrated] what was set out [to be proven]. 
◉ Hec quidem intelligenda sunt cum punctus visus non fuerit super perpendicularem cum ipso visu, tunc enim, cum infinite superficies possunt intelligi quarum quelibet ortogonalis super superficiem contingentem speculum, et omnes sint super illam perpendicularem, quedam illarum superficierum efficiet lineam communem lineam rectam, et non fiet reflexio nisi super eandem perpendicularem, et locus ymaginis centrum visus, et non videbitur punctus nisi qui fuerit in superficie visus. 
◉ These conclusions must be understood [to apply] when the visible point does not lie on the [same] normal as the center of sight, for in that case, since an infinite number of planes can all be imagined to lie on that normal such that each one is orthogonal to the plane tangent to the mirror, and since all of them lie on the normal, any one of those planes will form a rectilinear common section [with that tangent plane]; and reflection will only occur along the same normal, the center of sight [will constitute] the imagelocation, and no point will be seen unless it lies on the surface of the eye. 
◉ Quedam autem illarum superficierum efficit lineam communem circulum, et tunc puncta inter que et visum fuerit centrum circuli poterunt reflecti ad visum singula a duobus punctis circuli, cum a singulis ducantur linee facientes angulum cum superficie contingente quem per equalia dividat perpendicularis ducta ad centrum. Et hoc quidem dico de punctis que sunt in illa perpendiculari, et loca ymaginum erunt in centro circuli. Alia puncta illius perpendicularis non reflectentur ad visum preter punctum quod est in superficie visus, et illud per illam perpendicularem. 
◉ However, one of those planes forms a circular common section [with the mirror], and in that case, when the center of the mirror [D in figure 5.2.50a, p. 281] lies between the visible points [e.g., C] and the eye [E], each of those points can be reflected to the eye from two points [e.g., H and L] on the circle, since lines may be drawn from each of them to form an angle with the plane tangent [to the point of reflection such] that the diameter [HDL] drawn [from that point of reflection] to the center [of the circle] bisects [that angle]. I say this, of course, about points that lie on that normal, and their imagelocations lie at the center of sight.⁑ The other points on that normal [i.e., between D and E in figure 5.2.50a] will not be reflected to the eye except for the point that lies on the surface of the eye, and that one [is reflected] along the normal. 
◉ Cum autem fuerit linea communis sectio columpnaris, non poterunt puncta perpendicularis reflecti ab aliquibus punctis sectionis, cum forma accedens super perpendicularem reflectatur super perpendicularem, et in sectione unica sit perpendicularis, quare per hanc solam perpendicularem fiet reflexio, et solus punctus superficiei visus, et locus ymaginis centrum visus. 
◉ On the other hand, when the common section is a cylindric section, the points on the normal cannot be reflected from any [other] points on the section, because the form reaching along the normal must be reflected along the normal, and in [such] a section the normal is unique [as shown in book 4], so reflection will occur only along this normal, and only the point on the surface of the eye [will be so reflected], and the imagelocation will be at the center of sight. 
◉ Si vero fuerit centrum visus in centro circuli, reflectetur portio visus quam secant perpendiculares ducte a centro visus ad circulum a portione simili in circulo quam secant similiter eidem perpendiculares. Cum quelibet linea ducta a centro visus ad circulum sit perpendicularis, fiet reflexio per perpendicularem, et locus ymaginum centrum visus, quod est centrum circuli. 
◉ If, however, the center of sight lies at the center of the circle, the portion of the eye that the normals extending from the center of sight to the circle [on the mirror] cut off will be reflected from the corresponding portion of the circle [on the mirror] that the normals cut. Since any line extending from the center of sight to the circle is normal, reflection will occur along the normal, and the imagelocation will be [at] the center of sight, which is the center of the circle. 
◉ Amplius, fiat super punctum A angulus acutus quoque modo, qui sit FAG. Palam quoniam concurret FA cum GZ. Sit concursus in puncto Z, et fiat angulus CAG equalis angulo FAG. Concurret quidem AC cum GQ. Sit concursus in puncto C. Palam quoniam C refertur ad Z a puncto G, et etiam refertur ad Z a puncto A, et non ab alio puncto sectionis, quia non poterit reflecti nisi a termino perpendicularis, et una est in sectione illa perpendicularis, scilicet GA. 
◉ Now, at point A [figure 5.2.50, p. 280] form acute angle FAG of some kind. It is clear that FA will intersect GZ. Let the intersection be at point Z, and form angle CAG equal to angle FAG. AC will intersect GQ. Let the intersection be at point C. It is evident that [the form of point] C is reflected to [point] Z from point G, and it is also reflected to [point] Z from point A, but not from any other point on the section, because it cannot reflect except from the endpoint of the normal, and there is only one such normal in the section, namely, GA. 
◉ [PROPOSITIO 51] Amplius, sumptis duobus punctis in axe columpne, erit unum reflecti ad aliud ab uno circulo columpne toto, et loca ymaginum erit circulus quidam extra columpnam. 
◉ [PROPOSITION 51] Moreover, if two points are taken on the axis of the cylinder, [the form of] one will be reflected to the other from one full circle in the cylinder, and the imagelocations will lie on a given circle outside the cylinder. 
◉ Verbi gratia, sit EZ [FIGURE 5.2.51, p. 609] axis, T, H duo puncta sumpta in axe, AG, BD bases columpne. Dividatur TH per equalia in puncto Q, et fiat circulus cuius Q centrum, scilicet LM, qui erit equidistans basibus, eius dyameter LM, latera columpne BLA, DMG. Fiat etiam circulus KC cuius H centrum, CK dyameter, et ducantur linee TL, TM, HL, HM. 
◉ For example, Let EZ [figure 5.2.51, p. 282] be the axis, T and H two points selected on the axis, and AG and BD the bases of the cylinder. Bisect TH at point Q, and construct a circle with Q as its center, i.e., LM, that will be parallel to the [cylinder’s] bases, LM being its diameter, and BLA and DMG being sides of the cylinder. Also, construct circle KC with H as its center and CK its diameter, and draw lines TL, TM, HL, and HM. 
◉ Palam quoniam quatuor angulorum super Q quilibet est rectus, et TQ equalis QH, et QL equalis QM. Erunt illi trianguli similes, et anguli TLQ, QLH equales; similiter, anguli TMQ, QMH equales. Si igitur fuerit H centrum visus, reflectetur punctus T ad punctum H a puncto L, et similiter a puncto M. Si igitur moveatur triangulus TLH, immoto axe TH, describet punctus L circulum, et semper duo anguli TLQ, QLH manebunt equales, et semper in hoc motu reflectetur T ad H. 
◉ It is evident that each of the four angles at Q is a right angle, that TQ = QH, and that QL = QM. Those triangles [i.e., TLM and MLH] will be similar, so angles TLQ and QLH will be equal; likewise, angles TMQ and QMH will be equal. Therefore, if H is the center of sight, [the form of] point T will be reflected to point H from point L, and likewise from point M. Accordingly, if triangle TLH is rotated while axis TH remains stationary, point L will describe a circle, the two angles TLQ and QLH will remain constant throughout, and throughout this motion [the form of] T will be reflected to H. 
◉ Producatur autem linea CHK donec concurrat cum linea TL, et sit concursus F. Palam quoniam F erit locus ymaginis, et motu trianguli TLH, movebitur triangulus TFH, et hoc motu punctus F describet circulum extra columpnam. Et totus ille circulus erit locus ymaginum, et hoc est propositum. Idem erit probandi modus, sumptis quibuslibet duobus in axe punctis. 
◉ Now, draw line CHK until it intersects line TL, and let F be the [point of] intersection. It is evident that F will be the imagelocation, and as triangle TLH revolves, triangle TFH will revolve, and during this motion point F will describe a circle outside the cylinder. That entire circle will be the location for [all] images, and this is what was proposed.⁑ The same method of proof will apply for any two points [chosen] on the axis. 
◉ [PROPOSITIO 52] Amplius, punctorum extra perpendicularem visus sumptorum, quedam unicam habent ymaginem, quedam duas, quedam tres, quedam quatuor, non plures. 
◉ [PROPOSITION 52] Furthermore, some points that are selected outside the normal [extending] from the eye have one image, some [have] two, some [have] three, and some [have] four, but none [has] more [than four]. 
◉ Verbi gratia, sit A [FIGURE 5.2.52, p. 610] punctus visus extra perpendicularem visus, et fiat superficies transiens per A equidistans basibus speculi. Faciet quidem circulum in columpna. Sit centrum illius circuli H, et sumatur in superficie circuli aliud punctum, quod sit B, et ducantur dyametri AH, BH. 
◉ For instance, let A [figure 5.2.52, p. 283] be a visible point outside the normal [extending] from the center of sight, and construct a plane passing through A and parallel to the bases of the mirror. It will, of course, form a circle on the cylinder. Let H be the center of that circle, and choose another point B within the plane of the circle, and draw diameters AH and BH [B thus serving as a center of sight within this plane]. 
◉ Palam ex eis que dicta sunt in speculis spericis concavis quod ab uno puncto arcus quem intercipiunt hii duo dyametri potest A reflecti ad B forsitan a duobus, aut tribus, sed non a pluribus; ab arcu autem opposito non nisi ab uno puncto. Sit igitur quod A refertur ad B a tribus punctis arcus intercisi, et sint puncta illa G, D, E, et ducantur linee AG, HG, BG, AD, HD, BD, AE, HE, BE. 
◉ From what has been claimed about spherical concave mirrors [in proposition 49 above], it is clear that [the form of point] A may be reflected to [point] B from [at least] one point on the arc [passing through points E, D, and G] that subtends those two diameters, [or] perhaps from two or three, but from no more [than three]; and from the opposite arc [reflection can occur] from only one point. Accordingly, let [the form of point] A be reflected to [point] B from three points on the subtending arc [that passes through points E, D, and G], let those points be G, D, and E, and draw lines AG, HG, BG, AD, HD, BD, AE, HE, and BE. 
◉ Et a puncto A ducantur in eadem superficie tres linee equidistantes tribus dyametris HG, HD, HE, que sunt AK, AF, AN. Cum igitur AK sit equidistans HG, concurret BG cum AK. Concurrat in puncto K. Similiter BD concurret cum AF. Sit concursus in puncto F. Similiter BE cum AN. Sit concursus in puncto N. 
◉ From point A in the same plane draw three lines AK, AF, and AN parallel to the three diameters HG, HD, and HE [respectively]. Hence, since AK is parallel to HG, BG will intersect AK. Let it intersect at point K. Likewise, BD will intersect AF. Let the intersection be at point F. So, too, BE [will intersect] AN. Let the intersection be at point N. 
◉ Deinde a puncto H erigatur axis, que sit HU, et a puncto B perpendicularis super superficiem circuli. Erit quidem equidistans axi, que sit BT. Et sumatur in ea punctum quodcumque, quod sit T, et ducantur tres linee TK, TF, TN, et a tribus punctis G, D, E erigantur tres perpendiculares super superficiem circuli GM, DL, EQ. Erunt quidem equidistantes TB. EQ igitur erit in superficie trianguli TBN. Igitur EQ secabit TN. Secet in puncto Q. DL secet TF in puncto L; GM secet TK in puncto M. Et erunt hee tres perpendiculares linee longitudinis columpne. 
◉ Then, from point H erect axis HU, and from point B erect a line perpendicular to the plane of the circle. Let it be BT, and it will be parallel to the axis. Take some point T on it, draw the three lines TK, TF, and TN, and from the three points G, D, and E, erect three lines GM, DL, and EQ [respectively] perpendicular to the plane of the circle [each line thus being a line of longitude on the cylinder’s surface]. They will be parallel to TB. Hence, EQ will lie in the plane of triangle TBN. So EQ will intersect TN. Let it intersect at point Q. Let DL intersect TF at point L, and [let] GM intersect TK at point M. These three perpendiculars will constitute lines of longitude on the cylinder. 
◉ A puncto Q ducatur equidistans linee NA, que quidem concurret cum axe UH, quoniam erit equidistans EH. Sit concursus in puncto U, et ducatur linea TA, quam secabit QU, quoniam QU ducitur a latere trianguli equidistanter basi. Sit punctus sectionis I, et ducatur linea QA. 
◉ From point Q draw a line parallel to line NA, and it will intersect axis UH, because it will be parallel to EH [which is parallel to NA by construction]. Let the intersection be at point U, and draw line TA, which QU will intersect, because QU is drawn from one side of the triangle [i.e., TNA] parallel to the base [AN, which is parallel to HE by construction]. Let I be the point of intersection, and draw line QA. 
◉ Palam quoniam angulus BEH equalis est angulo ENA, et angulus HEA equalis angulo EAN, et angulus BEH equalis angulo HEA. Erit angulus EAN equalis angulo ENA, quare EN equalis EA. 
◉ It is clear that angle BEH = [alternate] angle ENA [since HE and AN are parallel, by construction], while angle HEA = [alternate] angle EAN [for the same reason], and angle [of reflection] BEH = angle [of incidence] HEA [by construction. Consequently] angle EAN = angle ENA, so EN = EA. 
◉ Et EQ perpendicularis. Erit triangulus QEA equalis triangulo QEN; erit QN equalis QA, et erit angulus QNA equalis angulo QAN. Sed angulus TQI equalis angulo QNA, et angulus IQA equalis angulo QAN. Erit angulus IQT equalis angulo IQA, quare A refertur ad T a puncto columpne quod est Q. 
◉ Moreover, EQ is perpendicular [to the plane of the circle. Accordingly] triangle QEA = triangle QEN [by Euclid, I.4]; [and so] QN = QA, and angle QNA = angle QAN [since triangle QNA is isosceles]. But angle TQI = [alternate] angle QNA [between parallels QI and NA], and angle IQA = [alternate] angle QAN [between parallels QI and NA]. [Hence], angle IQT = angle IQA, so [the form of point] A is reflected to [point] T from point Q on the cylinder.⁑ 
◉ Eodem modo probabitur quod refertur A ad T a punctis L, M, et ita a tribus punctis columpne ex eadem parte. 
◉ In the same way it will be demonstrated that [the form of point] A is reflected to [point] T from points L and M, and thus from three points on the same side of the cylinder.⁑ 
◉ Nec potest a pluribus, detur enim aliud. Ducto latere ab illo puncto, cadet in circulum quem habemus, et probabitur quod a puncto casus qui est in circulo poterit reflecti A ad T, replicata probatione, quod est impossibile. 
◉ Nor can it be reflected from more points, for let another [such point] be given. If a side [i.e., a line of longitude] is drawn from that point, it will fall on the circle that we have, and, by repeating the proof, it will be demonstrated [according to proposition 49 above] that from the point where the side falls to the circle it is impossible for [the form of point] A to be reflected to [point] T. 
◉ Ex arcu opposito circuli poterit reflecti A ad B ab uno puncto. Sit illud Z, et ducatur dyameter HZ, et ei equidistans AC. Et ducatur BZ, que concurrat cum AC in puncto C. Et erigatur perpendicularis OZ, que erit latus et equidistans TB, et ducatur TC, que secabitur a linea OZ. Sit sectio in puncto O. Probabitur modo predicto quod A refertur ad T a puncto O. Et si sumatur ex illa parte alius punctus columpne a quo possit reflecti, per replicationem probationis probabitur quod ab alio puncto circuli quam Z potest reflecti ex parte illa, quod est impossibile. 
◉ [But the form of point] A can be reflected to [point] B from one point on the opposite arc of the circle. Let Z [in figure 5.2.52, p. 283] be that point, and draw diameter HZ as well as line AC parallel to it. Then draw BZ, and let it intersect AC at point C. Erect perpendicular OZ, which will be a line of longitude and [thus] parallel to TB, and draw TC, which will be intersected by line OZ. Let the intersection be at point O. It will be proven according to the previous method that [the form of point] A is reflected to [point] T from point O.⁑ And if another point from which reflection can [supposedly] occur is chosen on that side of the cylinder, it will be demonstrated by repeating the proof [according to which the line of longitude is dropped from that point to the circle] that it is impossible for [the form of point A] to be reflected [to point B] from any point on that side of the circle other than Z [as demonstrated in proposition 38 above]. 
◉ Si ergo A ab uno puncto circuli refertur ad B ex aliqua parte, refertur ab uno columpne ex eadem; si a duobus, a duobus; si a tribus, a tribus; nec potest amplius; ab opposita parte ab uno circuli tantum, et ab uno columpne tantum. 
◉ Therefore, if [the form of point] A is reflected to [point] B from one point on a given side of the circle, it is reflected from one [point] on the same side of the cylinder; if [it is reflected] from two [points on the circle, it will be reflected] from two [points on the cylinder]; if [it is reflected] from three [points on the circle, it will be reflected] from three [points on the cylinder]; but [such reflection] can [occur] from no more [than three points on that side]; whereas on the opposite side [it can occur] from only one point on the circle and [thus] from only one point on the cylinder. 
◉ Item, TB equidistans UH, nec potest sumi superficies equalis in qua sit T cum UH preter superficiem TBUH. Similiter, non potest sumi superficies in qua sit A cum UH preter superficiem AUH, que est perpendicularis. T igitur non est in eadem superficie perpendiculari cum A, nec in eodem circulo, nec est in axe, quia est in linea ei equidistante. Superficies igitur in qua A refertur ad T est sectio columpnaris. 
◉ Furthermore, TB is parallel to UH, and there can be no plane chosen, other than plane TBUH, in which T lies with UH. Likewise, there can be no plane other than AUH in which A lies with UH, and it is perpendicular [to the plane of the basecircle]. Hence, T does not lie in the same perpendicular plane with A, nor on the same circle [forming its plane], nor is it on the axis, because it lies on a line parallel to it. Accordingly, the plane in which [the form of point] A is reflected to [point] T constitutes a cylindric section [because it is oblique]. 
◉ Verum, producta TA ultra T et A ex utraque parte, et sit RP. Cum quatuor sint superficies reflexionis, quia a quatuor punctis, et in qualibet sint duo puncta T, A, erit RP communis quatuor superficiebus reflexionis. Et quelibet harum superficierum secat superficiem contingentem speculum in puncto super suam lineam communem, non super eandem. Linea RP perpendicularis est super unam linearum quatuor communium, non super duas, esset enim perpendicularis super superficiem contingentem, et ita perveniret ad axem. Sunt igitur diverse perpendiculares a puncto T ad has quatuor lineas communes, nec est nisi una tantum que transeat per A. 
◉ Now, let TA be extended beyond T and A on both sides to form RP. Since there are four planes of reflection, because [reflection occurs] from four points, and since the two points T and A lie in each of them, RP will be common to the four planes of reflection. Moreover, each of these planes cuts the plane tangent to the mirror at a point on its own common section [with the mirror], but not on the same common section [as any other plane of reflection].⁑ Line RP is normal to one of the four common sections, but not to two, for if it were perpendicular to the tangent plane [common to two or more common sections], it would thus reach the axis. Hence, the normals [dropped] from point T to these four common sections are distinct [from one another], and there is only one that passes through A.⁑ 
◉ Et perpendicularis aut erit equidistans linee reflexionis, aut concurret cum ea ultra speculum, vel intra. Si fuerit equidistans, erit locus ymaginis punctus reflexionis, ut probatum est, et cum quatuor sint reflexionis puncta, erunt quatuor ymagines. Si concurrit, cum quatuor sint perpendiculares, quatuor erunt concursus, et quatuor ymagines. [2.519] Amplius, datis puncto viso et puncto visus, erit invenire punctum reflexionis. Verbi gratia, sit A punctus visus. Fiat superficies secans columpnam equidistans basi transiens per A, et faciet circulum. B aut est in superficie huius circuli, aut non. Si fuerit, inveniemus punctum reflexionis in illo circulo sicut dictum est in sperico concavo. Si non fuerit, ducatur a puncto B perpendicularis super superficiem huius circuli, et replicetur supradicta probatio, et invenietur punctus reflexionis. Duplici autem visu adhibito, una ymago in veritate efficientur due, sed contigue vel admixte, unde videbuntur una. 
◉ Moreover, the normal [within any of the planes of reflection] will either be parallel to the line of reflection or will intersect it beyond or inside the mirror. If it is parallel, the point of reflection will be the image location, as has been demonstrated [in proposition 32], and since there are four points of reflection, there will be four images. If it intersects, then, since there are four normals there will be four intersections and four images. 
◉ In speculis piramidalibus concavis, linea communis superficiei reflexionis et superficiei speculi aut erit linea longitudinis speculi, aut erit sectio piramidalis. Si fuerit linea longitudinis, erunt loca ymaginum in ipso speculo. Si fuerit sectio piramidalis, erunt loca ymaginum aliquando citra visum, aliquando in visu, aliquando inter visum et speculum, aliquando ultra speculum, sicut ostensum est in speculo columpnari concavo. 
◉ In concave conical mirrors, the common section of the plane of reflection and the surface of the mirror will be a line of longitude along the mirror, or it will be a conic section. If it is a line of longitude, the imagelocations will lie in [i.e., behind] the mirror[‘s reflecting surface].⁑ If it is a conic section, the imagelocations will sometimes lie beyond the center of sight, sometimes at the center of sight [itself], sometimes between the center of sight and the mirror, and sometimes behind the mirror, just as was shown in the case of the concave cylindrical mirror. 
◉ Amplius, si in perpendiculari ducta a centro visus ad superficiem contingentem piramidem sumatur punctus corporeus inter visum et speculum, non reflectetur forma eius ad visum per perpendicularem, quoniam punctus ille occultabit terminum perpendicularis, et ob hoc non reflectetur ab eo. Si autem nullus fuerit punctus in perpendiculari illa, reflectetur quidem ad visum per hanc perpendicularem punctum visus, quod iterum secat perpendicularis ex eo, et ille solus. 
◉ Furthermore, if a physical spot is taken on the normal extending from the center of sight to the plane tangent to the mirror [and if it lies] between the center of sight and the mirror, its form will not be reflected to the center of sight along the normal, because that spot will block the endpoint of the normal [at the eye], and for that reason it will not be reflected from it. However, if there is no such [physical] spot on that normal, [the form of] a visible point will be reflected to the eye along this normal, that point, and that point only, being the one on [the surface of the eye] that intersects the normal. 
◉ Verum, visu existente in hac perpendiculari et in axe, efficietur circulus ad cuius quodlibet punctum linea ducta a visu erit perpendicularis super superficiem contingentem, unde a quolibet puncto illius circuli fieri poterit reflexio ad visum per perpendicularem. Et fiet reflexio partis visus quam secant due perpendiculares maiorem angulum in eo continentes. 
◉ On the other hand, if the center of sight lies on that normal as well as on the axis, it will form a circle, and the line drawn to any point on it from the center of sight will be normal to the plane tangent [to the mirror at that point], so from any point on that circle reflection can occur to the eye along the normal. And the portion of the eye that the two normals cut off to form the greatest [visible] angle on it will be reflected. 
◉ Si vero inter visum et speculum fuerit axis, non fiet ad ipsum reflexio per perpendicularem nisi puncti eius quem secat perpendicularis. 
◉ If, however, the axis lies between the center of sight and the mirror, there will be no reflection to it along the normal except for that point on it that the normal intersects.⁑ 
◉ [PROPOSITIO 53] Amplius, existente visu et puncto viso in axe, erit reflecti unum ad aliud. 
◉ [PROPOSITION 53] Now, if the center of sight and the visible point lie on the axis, [the form of] the latter will be reflected to the former. 
◉ Verbi gratia, sit H [FIGURE 5.2.53, p. 611] centrum visus, T punctus visus. Fiat superficies secans piramidem transiens super axis longitudinem, que sit ABGH, AH axis, AB, AG latera piramidis. A puncto T ducatur perpendicularis super lineam AB, que sit TQ, et producatur usque QL. Sit equalis QT. Et a puncto H ducatur linea ad punctum L, que secabit lineam longitudinis que est AB. Secet in puncto B, et a puncto B ducatur equidistans linee TQ, que necessario perveniet ad axem. Perveniat in puncto D, et ducatur linea TB. 
◉ For instance, let H [figure 5.2.53, p. 286] be the center of sight, and T the visible point. Construct a plane that cuts the cone along the length of the axis, and let it be ABGH, with AH the axis and AB and AG edges of the cone. From point T draw TQ perpendicular to line AB, and extend it to QL. Let [QL] = QT. Then, from point H draw a line to point L, and it will intersect line of longitude AB. Let it intersect at point B, and from point B draw a line parallel to line TQ, which will necessarily reach the axis. Let it reach [the axis] at point D, and draw line TB. 
◉ Palam, cum TQ sit perpendicularis super AB, et TQ equalis QL, erit triangulus BTQ equalis triangulo BQL, et erit angulus QLB equalis angulo QTB. Sed angulus QTB equalis est angulo TBD, et angulus DBH equalis est angulo QLB. Igitur angulus TBD equalis est angulo DBH, et ita T refertur ad H a puncto B, et locus ymaginis L. 
◉ Since TQ is perpendicular to AB, and since TQ = QL, it is clear that triangle BTQ = triangle BQL, and angle QLB = angle QTB. But angle QTB = [alternate] angle TBD, and angle DBH = [alternate] angle QLB. Therefore, angle TBD = angle DBH, and so [the form of point] T is reflected to H from point B, and L is the imagelocation. 
◉ Igitur, moto triangulo TLH, describet punctus B circulum in piramide, et a quolibet puncto illius circuli reflectetur T ad H. L vero extra circulum describet circulum qui totus erit locus ymaginis puncti T. 
◉ Accordingly, if triangle TLH is rotated [about axis TH], point B will describe a circle on the cone, and from any point on that circle [the form of point] T will be reflected to [point] H. Meanwhile, outside this circle, L will describe a circle that will constitute in its entirety the imagelocation for point T. 
◉ [PROPOSITIO 54] Amplius, sumptis duobus punctis extra perpendicularem visus et extra axem in hoc speculo, scilicet Z,E [FIGURE 5.2.54, p. 612], fiat superficies equidistans basi super Z. Faciet circulum in speculo. E aut erit in hoc circulo, aut in alia superficie. 
◉ [PROPOSITION 54] Now, in this [sort of] mirror, having selected two points, i.e., Z and E [figure 5.2.54, p. 287] outside the normal [extending] from the center of sight and outside the axis, construct a plane on [point] Z parallel to the base [of the cone]. It will produce a circle in the mirror. E will lie in this circle or in another plane. 
◉ Sit in superficie illius circuli, et ducatur linea EZ. Palam quoniam Z refertur ad E a circulo illo ex una parte aut ab uno puncto, aut a duobus, aut a tribus; ex alia ab uno. 
◉ Let it lie in the plane of that circle, and draw line EZ. It is evident [from proposition 49, case 3 above] that [the form of point] Z is reflected to E on one side of that circle from one point, or from two, or from three; and on the other side [it is reflected] from one [point]. 
◉ Sumatur punctus circuli a quo refertur ad ipsum, et sit H, centrum circuli T. Et ducantur linee ZH, EH. Et dyameter TH dividet quidem angulum illum per equalia, et secabit lineam EZ. Secet in puncto Q, et sit A conus piramidis, AH linea longitudinis. 
◉ Take a point on the circle from which [the form of point Z] is reflected [to E], let it be H, and [let] T [be] the center of the circle. Draw lines ZH and EH. Diameter TH will bisect the angle [formed by them], and it will intersect line EZ. Let it intersect at point Q, let A be the vertex of the cone and AH a line of longitude. 
◉ A puncto Q ducatur linea cadens perpendiculariter super lineam AH, que sit QM, que quidem perveniat ad axem, qui est AT. Cadat in ipsum in puncto D, et ducantur linee ZM, EM. A puncto Z ducatur in superficie circuli linea equidistans linee QH, que sit ZL. Concurrat quidem EH cum illa. Sit concursus in puncto L, et a puncto H ducatur perpendicularis super LZ, que sit HC. 
◉ From point Q draw line QM falling orthogonally to line AH, and let it reach the axis [AT]. Let it fall at point D on the axis, and draw lines ZM and EM. From point Z in the plane of the circle draw line ZL parallel to line QH. Let EH intersect it. Let the intersection be at point L, and from point H draw HC perpendicular to LZ. 
◉ Deinde in superficie trianguli EMZ ducatur linea equidistans linee QM, que sit ZO. Concurrat EM cum ea in puncto O, et ducatur linea LO. Et a puncto C ducatur equidistans LO, que sit CN, et ducatur linea NM. 
◉ Then, in the plane of triangle EMZ draw line ZO parallel to line QM. Let EM intersect it at point O, and draw line LO. From point C draw CN parallel to LO, and draw line NM. 
◉ Palam quoniam angulus EHQ equalis est angulo QHZ et angulo HLZ, et angulus QHZ equalis est angulo HZL. Erit HL equalis HZ, et HC perpendicularis super LZ. Erit triangulus LCH equalis triangulo CHZ, et erit LC equalis CZ. 
◉ It is clear [by construction] that angle EHQ = angle QHZ as well as [alternate] angle HLZ [since ZL is parallel to QH by construction], and angle QHZ = [alternate] angle HZL. [Thus] HL = HZ, and HC is perpendicular to LZ [by construction. Since, therefore, HC is common to triangles LCH and CHZ] triangle LCH = triangle CHZ [by Euclid, I.26], and [so] LC = CZ. 
◉ Et CN equidistans OL; erit proportio LC ad CZ sicut ON ad NZ, quare ON equalis NZ. Item, cum OZ sit equidistans QM, erit superficies ZLO equidistans superficiei QMH. Et superficies EOL secat illas duas super lineas communes que quidem erunt equidistantes, scilicet MH, OL, quare HM, CN equidistantes. Et quoniam HC cadit inter LZ, HQ equidistantes, et est perpendicularis super LZ, erit perpendicularis super HQ, quare CH erit contingens circulo. [2.535] Igitur superficies AHC est superficies contingens piramidem. In hac superficie est CN et NM, et super hanc superficiem est perpendicularis linea DM. Igitur perpendicularis est super lineam NM, quare NM perpendicularis super OZ, et ON equalis NZ. Erit MO equalis MZ, et proportio EM ad MO sicut EM ad MZ. 
◉ CN is parallel to OL [by construction, so, by Euclid, VI.2] LC:CZ = ON:NZ, so ON = NZ. Furthermore, since OZ is parallel to QM [by construction], plane ZLO will be parallel to plane QMH. Plane EOL intersects these two [planes] along [rectilinear] common sections, i.e., MH and OL, that will be parallel, so HM and CN are parallel [because CN is parallel to OL, by construction]. And because HC falls between parallels LZ and HQ, and since it is perpendicular to LZ, it will be perpendicular to HQ, so CH will be tangent to the circle. 
◉ Sed EM ad MO sicut EH ad HL, et EH ad HL sicut EH ad HZ, et EH ad HZ sicut EQ ad QZ. Igitur EM ad MZ sicut EQ ad QZ, quare angulus EMQ equalis angulo QMZ, quare Z refertur ad E a puncto M. Si ergo Z refertur ad E a puncto circuli H, refertur ad ipsum a puncto piramidis M. Et si a duobus circuli, a duobus piramidis; si a tribus, a tribus; si a pluribus, a pluribus. Eodem modo ex alia parte circuli fiet probatio quod ab uno piramidis sicut ab uno circuli. 
◉ But EM:MO = EH:HL [by Euclid, VI.2], EH:HL = EH:HZ [by Euclid, V.7, because HL = HZ by previous conclusions], and EH:HZ = EQ:QZ [because HQ bisects angle EHZ]. Thus, EM:MZ = EQ:QZ [since EM:MZ = EM:MO = EH:HL = EH:HZ], so [because the respective sides of triangles EMQ and ZMQ are proportional, making the two triangles similar] angle EMQ = angle QMZ, [and] so [the form of point] Z is reflected to E from point M. Hence, if [the form of point] Z is reflected to E from point H on the circle, it is reflected to the same point from point M on the cone. And if [it is reflected] from two [points] on the circle, [it is reflected] from two [points] on the cone; if [it is reflected] from three [points on the circle, it is reflected] from three [points on the cone]; and if [it is reflected] from more [points on the circle, it is reflected] from more [points on the cone].⁑ On the other side of the circle, the proof that [reflection occurs] from one point on the cone just as from one [point] on the circle will be constructed in the same way. 
◉ Si vero E non fuerit in circulo equidistante basi transeunte super Z, erit E [FIGURE 5.2.54a, p. 613] quidem supra aut infra. Sit supra, quia utrobique eadem est probatio. Ducatur linea AE donec contingat superficiem illius circuli, et sit punctus contactus H, Q centrum circuli. Palam quoniam H potest reflecti ad Z ab aliquo puncto circuli. Sit illud T, et ducatur dyameter QT. Et linea HZ secabit hunc dyametrum in puncto quod sit N. Et ducatur EZ et linea longitudinis AT. 
◉ However, if E does not lie on the circle that passes through Z parallel to the base [of the cone], E will lie above or below it. Let it lie above, since the same proof applies to both cases. Draw line AE [figure 5.2.54a, p. 288] until it touches the plane of that circle, and let H be the point of contact, Q being the center of the circle. It is obvious that [the form of point] H can be reflected to Z from some point on the circle. Let it be T, and draw diameter QT [normal to the point of reflection]. Line HZ will intersect this diameter at point N. Draw [line] EZ and line of longitude AT. 
◉ Palam, cum punctus Z sit ex una parte dyametri QT, ex alia E, linea EZ secabit superficiem AQT. Secet in puncto O, et a puncto O ducatur perpendicularis super lineam AT, que sit OC, que necessario cadet super axem. Cadat in puncto D, et ducantur linee EC, ZC. Dico quoniam E refertur ad Z a puncto C. 
◉ Since point Z lies on one side of diameter QT, and since [point] E lies on the other, it is evident that line EZ will intersect plane AQT. Let it intersect at point O, and from point O draw [line] OC perpendicular to line AT, and it will necessarily fall upon the axis. Let it fall at point D, and draw lines EC and ZC. I say that [the form of point] E is reflected to Z from point C. 
◉ Probatio: ducatur a puncto Z linea equidistans QT, que sit ZF, et producatur linea HT donec concurrat cum illa. Sit concursus in puncto F. Similiter, a puncto Z ducatur equidistans linee OC, que sit ZK, et producatur linea EC donec concurrat cum illa. Sit concursus in puncto K. 
◉ [Here is] the proof. From point Z draw line ZF parallel to [line] QT, and extend line HT until it intersects it. Let the intersection be at point F. Likewise, from point Z draw [line] ZK parallel to line OC, and extend line EC until it intersects it. Let the intersection be at point K. 
◉ Palam, cum linea ZF sit equidistans QT, et ZK equidistans OC, erit superficies ZKF equidistans superficiei OCT que est superficies AQT. Et superficies HFK secat has duas superficies super lineas CT, KF. Igitur CT, KF sunt equidistantes. 
◉ Since line ZF is parallel to [line] QT [by construction], while [line] ZK is parallel to [line] OC [by construction], it is clear that plane ZKF will be parallel to plane OCT, which lies within plane AQT [since O is where HK intersects plane AQT]. Plane HFK intersects these two planes along lines CT and KF. Hence, CT and KF are parallel. 
◉ Ducatur a puncto T perpendicularis super lineam ZF, que sit TP. Palam, cum cadat inter duas equidistantes, erit equidistans linee NZ, et ita erit contingens circulo. Igitur superficies ATP contingit piramidem super lineam AT, et linea OC est perpendicularis super hanc superficiem. Superficies igitur ATQ erit ortogonalis super superficiem ATP, et superficies ATP secat duas superficies ATQ, ZKF, que sunt equidistantes. Igitur linee communes sectionum sunt equidistantes, una harum linearum est CT, alia sit PI. Sed iam patet quod CT est equidistans KF. Igitur PI est equidistans KF. 
◉ From point T draw [line] TP perpendicular to line ZF. Since it falls between two parallels [i.e., TQ and ZF], it is obvious that it will be parallel to line NZ, and so it will be tangent to the circle [at point T].⁑ Hence, plane ATP is tangent to the cone along line [of longitude] AT, and line OC is perpendicular to this plane. Accordingly, plane ATQ will be orthogonal to plane ATP, and plane ATP intersects the two planes ATQ and ZKF, which are parallel. Thus, the common sections, one being CT, the other PI, are parallel. But it has already been shown that CT is parallel to KF. Hence, PI is parallel to KF. 
◉ Sed planum est quod angulus NTZ equalis est angulo TZF, et angulus HTN equalis angulo TFZ, et TP perpendicularis. Erit FP equalis PZ. Sed proportio FP ad PZ sicut KI ad IZ; erit KI equalis IZ. 
◉ But it is evident that [since ZT falls between parallels NT and ZF] angle NTZ = [alternate] angle TZF, whereas angle HTN = [alternate] angle TFZ, and TP is perpendicular [to ZF. Therefore, by Euclid, I.26] FP = PZ. But FP:PZ = KI:IZ, [so, by Euclid, VI.2] KI = IZ. 
◉ Ducta autem linea CI, cum superficies ATPI sit ortogonalis super superficiem ZKF, erit CI ortogonalis super ZK, et erit angulus CKZ equalis angulo KZC. Sed angulus ECO equalis est angulo CKZ, et angulus OCZ equalis est angulo CZK, quare angulus ECO equalis est angulo OCZ. Et ita E refertur ad Z a puncto C, quod est propositum. 
◉ Now, if line CI is drawn, then, since plane ATPI is perpendicular to plane ZKF [which is parallel to plane ATQ], CI will be perpendicular to ZK, and angle CKZ = angle KZC. But angle ECO = angle CKZ [by Euclid, I.29, because OC and KZ are parallel, and EC cuts them both], and angle OCZ = [alternate] angle CZK, so angle ECO = angle OCZ. Hence, [the form of point] E is reflected to [point] Z from point C, which is what was proposed. 
◉ Si autem sumatur aliud punctum in circulo a quo H reflectatur ad Z, probabitur quod ab alio puncto piramidis quam C refertur E ad Z. Et si reflectatur H ad Z a tribus punctis circuli, reflectetur E ad Z a tribus piramidis; si a quatuor, a quatuor. 
◉ Moreover, if another point is taken on the circle from which [the form of point] H is reflected to [point] Z, it will be demonstrated that [the form of point] E is reflected to Z from some point other than C on the cone. And if [the form of point] H is reflected to [point] Z from three points on the circle, [the form of point] E will be reflected to [point] Z from three [points] on the cone; if [it is reflected] from four [points on the circle, it will be reflected] from four [points on the cone]. 
◉ Punctum autem reflexionis a quo E refertur ad Z facile est invenire, invento puncto circuli a quo punctus H refertur ad Z, et erit inventio modo predicto. 
◉ Furthermore, the point of reflection from which [the form of point] E is reflected to [point] Z is easy to find when the point on the circle from which [the form of] point H is reflected to [point] Z is found, and it will be found in the preceding way. 
◉ Si vero dicatur quod a pluribus punctis piramidis quam quatuor possit punctus E reflecti ad Z, per replicationem predicte probationis poterit ostendi quod punctus H refertur ad Z a pluribus punctis circuli quam quatuor, et ubi accidet punctum H reflecti ad Z ab aliquot punctis circuli vel ab uno tantum, accidet punctum E reflecti ad Z a totidem punctis piramidis aut ab uno tantum, et econverso. Quod si dicatur contrarium, poterit improbari predicto modo. 
◉ If, however, it were claimed that [the form of] point E could be reflected to [point] Z from more than four points on the cone, it would be possible by repeating the earlier proof to show that [the form of] point H is reflected to [point] Z from more than four points on the circle, and in the case where [the form of] point E will happen to be reflected to [point] Z from however many points on the circle, or from one only, [the form of] point E will happen to be reflected to [point] Z from that many points on the cone, or from one only, and viceversa. But if [this] contrary claim is made, it can be disproven in the preceding way [i.e., according to proposition 49]. 
◉ Palam ergo quod punctorum quedam habent unicam ymaginem, quedam duas, quedam tres, quedam quatuor, sed non possibile quod plures. Verum adhibito speculo duplici visu eiusdem ymaginis diversa erunt loca, que diversitas, propter sui imperceptibilitatem, non inducit errorem. 
◉ Hence, it is clear that some of the points [that are reflected] have a single image, some [have] two, some three, and some four, but no [more] than four are possible. In addition, when the mirror is exposed to both eyes, the same image will have different locations, but, because of its imperceptibility, this difference [in location] does not cause an error [in visual perception]. 