Liber sextus
◉Liber iste in novem partes partitur. Pars prima, titulus libri; secunda, quoniam error accidit visui propter reflexionem; tertia, in errore evenienti in speculis planis; quarta, in errore qui oritur in speculis spericis exterioribus; quinta in speculis columnaribus exterioribus; sexta, in piramidalibus exterioribus; septima, in spericis concavis; octava, in columpnis concavis; nona, in piramidalibus concavis. 
◉This book is divided into nine chapters. The first chapter [describes] the basic purport of the book; the second [explains] that error occurs in sight because of reflection; the third [focuses] on error that arises in plane mirrors; the fourth [focuses] on error that originates in convex spherical mirrors; the fifth [focuses] on convex cylindrical mirrors; the sixth [focuses] on convex conical [mirrors]; the seventh [focuses] on concave spherical [mirrors]; the eighth [focuses] on concave cylindrical [mirrors]; the ninth [focuses] on concave conical [mirrors]. 
◉Pars prima 
◉CHAPTER 1 
◉ Patuit ex libris superioribus modus adquisitionis formarum in speculis per visum, situs linearum reflexionis vel accessus, situs ymaginum et loca ipsarum. Verum per reflexionem non semper comprehenditur forme veritas. In concavis enim speculis apparet ymago faciei distorta, et occultatur visui dispositio ipsius vera, unde planum errorem incidere in comprehensione formarum per reflexionem. Huius erroris modum et modi causam propositum est in libro presenti explanare et secundum diversitates speculorum disquirere varietates errorum. 
◉ It was shown in the preceding books how forms are apprehended in mirrors by the visual faculty, how the lines of reflection or incidence are disposed, [and] how images are disposed and where they are located. However, the form is not always perceived as it actually exists by means of reflection. For in concave mirrors the image of [one’s] face appears distorted, and its proper disposition is obscured from sight, so it is obvious that error occurs in the perception of forms through reflection. In the present book it is [our] purpose to explain how this error occurs and the reason for it, as well as to discuss the different types of errors due to the different types of mirrors. 
◉Pars secunda 
◉CHAPTER 2 
◉ Comprehensionem formarum in visu directo liber secundus docuit, et singula que propter egressum a temperantia in visu illo errorem inducunt liber tertius diligenter exposuit. Fit autem comprehensio formarum per reflexionem sicut et directe, et quorum fit adquisitio in directione fit etiam in reflexione, utpote lucis, coloris, figure, magnitudinis, distantie, et similium. 
◉ The second book showed how forms are perceived in direct vision, and the third book carefully analyzed the particular factors that lead to error in that [kind of] vision when the [conditions for proper vision] exceed or fall short of the [appropriate] threshold. The perception of forms by means of a reflection [of rays] occurs in the same way [as it does] in direct vision, and [so] the things that are apprehended in direct vision are also apprehended in reflected vision—such things as light, color, shape, size, distance, and the like [i.e., the full range of visible intentions]. 
◉ Et quemadmodum in directione rerum prefixarum et cognitarum ad alia fit collatio, et inde oritur coniecturatio, et sumitur iudicium in anima, similiter accidit in reflexione. Unde quecumque temperamentum egressa in visu directo errorem efficiunt, in reflexione similiter inducunt. Et secundum singula maior accidit error in reflexione propter lucem debilem quam debilitat ipsa reflexio. 
◉ Moreover, just as happens in the direct visual apprehension of things [whose forms] are already ensconced [in the soul] and known, so in reflected vision there is a correlation [of the form] to something else [like it] so that a conclusion is drawn and a judgment is made in the soul. Hence, any excesses or defects in the threshold conditions [for proper sight that] cause an error in direct vision likewise cause [an error] in reflected vision. And according to each case [of excess or defect in the threshold condition], the error is magnified in reflected vision because of the diminished light that results from the weakening caused by the actual reflection. 
◉ Ut autem generaliter loquamur, non potest in reflexione comprehendi veritas forme sicut potest in directione propter triplex impedimentum reflexioni speciale. Primum est quoniam in reflexione apparet rei forma pre oculis visui opposita, cum non sit re vera. Secundum quoniam lux et color corporis visi miscentur cum colore speculi, quam mixturam visus percipit non verum rei vise colorem vel lucem. Tertium quoniam ipsa reflexio, ut in superioribus est assignatum, lucem et colorem debilitat, quare in reflexione latebit visum veritas lucis et coloris plus quam in directione. 
◉ Furthermore, to generalize, we should say that the proper disposition of the form cannot be perceived in reflected vision as it can be in direct vision because of a threefold constraint specific to reflection. The first is that in reflection the form of the object appears to the viewer to lie directly in front of the eyes when this is not actually the case. The second [is] that the light and color in the visible object are mingled with the color of the mirror, and the visual faculty perceives that mingled [color] rather than the actual color or light belonging to the visible object. The third is that, as has been pointed out earlier [in book 4], reflection itself weakens light and color, so the actual light and color will be less clearly seen in reflected vision than in direct vision. 
◉ Amplius superiora docuerunt quoniam quantitas temperamenti eorum que in visu directo errorem inducunt fortitudinem lucis et coloris respicit, fortiore enim luce vel colore erit maior, debiliore minor. Cum autem per reflexionem debilitentur lux et color, erit latitudo temperamenti singulorum errorem inducentium minor in reflexione quam in directione, et temperantie diminuta latitudo pluralitatem erroris inducit. Preterea quedam minutie corporum comprehendi poterunt per directionem que nullatenus comprehensibiles sunt per reflexionem. Palam ergo quod directionem superat reflexio in maioritate errorum et numero. 
◉ In addition, earlier discussions showed that the range of the [limits of the] threshold conditions [whose excess or defect] leads to error depends on the intensity of the light and color, for that range will be greater in stronger light or color [and] less in weaker [light or color]. And, since light and color will be weakened by reflection, the range of the [limits of the] threshold conditions [whose excess or defect] leads to particular kinds of error will be less in reflected vision than in direct vision, and the shortening of that range leads to an increase in the number of errors. Besides, certain tiny features of objects can be perceived through direct vision that are in no way perceptible through reflected vision. It is therefore evident that reflected vision exceeds direct vision in the degree and number of errors. 
◉Pars tertia 
◉CHAPTER 3 
◉ In singulis speculis erronea formarum accidit comprehensio, sed iuxta varietatem speculorum fit varietas errorum. In speculis planis minor accidit error quam in aliis. In hiis etenim comprehenditur veritas figure, situs, et quantitatis, sicut in directione, quod per probationem patebit. 
◉ In each kind of mirror a misperception of forms occurs, but the variety of errors [that occur] depends on the variety of mirrors [in which the forms are perceived]. In plane mirrors less error occurs than in the others. For in these [kinds of mirrors] the proper shape, spatial disposition and size [of the object] are perceived, just as [they are] in direct vision, which will be shown by [the following] demonstration. 
◉ [PROPOSITIO 1] Proponatur speculum planum [FIGURE 6.3.1, p. 304], et sit AB linea in superficie illius speculi communis superficiei speculi et superficiei orthogonali super superficiem speculi. Sint H, Z duo puncta in superficie illa orthogonali, E centrum visus, et a puncto H ducatur perpendicularis super superficiem speculi, que sit HL. Et producatur ut LG sit equalis LH. Similiter, producatur perpendicularis ZF ut DF sit equalis FZ. 
◉ [PROPOSITION 1] Imagine a plane mirror, and let line AB [in figure 6.3.1, p. 97] on that mirror’s surface be the common section of the mirror’s surface and a plane perpendicular to the mirror’s surface. Let H and Z be two points in that perpendicular plane, [let] E [be] the center of sight, and draw perpendicular HL from point H to the mirror’s surface. Extend it so that LG = LH. Likewise, extend perpendicular ZF so that DF = FZ. 
◉ Planum ex superioribus quoniam H refertur ad E a puncto speculi, et locus ymaginis ipsius est G, tantum distans a superficie speculi quantum H. Similiter, Z refertur ad E, et locus ymaginis est D. 
◉ It is clear from earlier discussions [i.e., book 5, proposition 1, in Smith, Alhacen on the Principles, 399] that [the form of point] H is reflected to [point] E from a point on the mirror, and its imagelocation G lies as far from the mirror’s surface [below it] as H [lies above it]. By the same token, [the form of point] Z is reflected to [point] E, and its imagelocation is D. 
◉ Ducta autem linea ZH, et similiter linea GD, quodcumque punctum linee ZH refertur ad E. Locus ymaginis eius est tantum distans a superficie speculi quantum ipse punctus, et ita quilibet punctus linee ZH tantum videtur distare quantum distabit. Unde, si linea ZH fuerit recta, erit linea DG recta. Si fuerit arcus, erit DG arcus et eiusdem curvitatis, quare linea ZH apparebit eiusdem quantitatis, eiusdem figure cuius fuerit, quod est propositum. 
◉ Now when line ZH is drawn, and likewise line GD, [the form of] any point on line ZH is reflected to [point] E. Its imagelocation lies the same distance from the mirror’s surface as the point itself, and so any point on line ZH appears to lie the same distance [from the mirror’s surface] as it will [actually] lie [from that surface]. Hence, if line ZH is straight, line DG will be straight. If it is curved, DG will be an arc of the same curvature, so line ZH will appear the same size and shape as it is, which is what was set out [to be proven]. 
◉ Verum, si in punctis linee ZH fuerit varietas colorum minutim variata, forsan non discerneretur variatio; sed una pretendetur visui coloris confusio. Unde erit error in luce et in colore, et hoc in numero propter reflexionem. Illa etenim colorum et lucium varietas forsan comprehendi posset directe, sed egressus est color a temperantia respectu reflexionis, non respectu directionis. Similiter, minutie occultantur aut confunduntur in reflexione que discerni possent in directione. 
◉ However, if there are various colors that are only slightly different from one another at points along line ZH, the variation [among them] may not be perceived; instead, a single blend of color will be presented to sight. Hence, because of reflection there will be an error involving light and color, and in addition [an error] concerning number. For that difference among the colors and lights might be perceptible in direct vision, but the [perceptibility of the] color has exceeded the threshold with respect to reflected vision, although not with respect to direct vision. Likewise, tiny features that could be discerned in direct vision are either hidden or confused in reflected vision. 
◉ Et propter debilitationem lucis vel coloris ex reflexione accidit error in longitudine qui quidem non accideret directe. 
◉ Moreover, because of the weakening of light or color by reflection, an error arises in [perception] of distance that would not arise in direct vision.⁑ 
◉ In situ manifeste accidit error ex sola reflexione, in ymagine enim sinistra comprehendimus ea que in corpore viso, si esset in loco ymaginis, dextra videremus. Cum enim aliquid alii opponitur, contrarius est eis adinvicem situs, quod enim uni fuerit dextrum alii erit sinistrum. Igitur quod rei vise dextrum est ymagini sinistrum, et sinistrum in ymagine dextrum erit videnti, sed comprehenditur in ymagine sinistrum. 
◉ In the case of spatial disposition an error clearly arises from reflection alone, for in the image we perceive things on the lefthand side of the visible object that we would see on the righthand side if the object were [actually placed in front of us] at the imagelocation. For, when something faces something else, its corresponding spatial disposition is opposite because what is the righthand side of the one will be the lefthand side of the other. Accordingly, the righthand side of the visible object is the lefthand side of the image, whereas the lefthand side of the image will be its righthand side to the viewer, but it is perceived on the lefthand side of the image.⁑ 
◉ Et generaliter in modo lucis, vel coloris, vel situs error semper accidit ex sola reflexione. In hiis et in aliis que errorem inducunt directe inducunt similiter in reflexione, et facilius, quoniam temperamentum singulorum minus est visui reflexo quam in directo. Horum omnium unum apponatur exemplum, et idem in ceteris intelligatur. 
◉ Overall, in the case of light, color, or spatial disposition, error invariably arises from the very reflection itself. In these cases, as well as in others, the things that lead to error in direct vision likewise lead to error in reflected vision, and more easily because the [range of] threshold conditions for each is smaller in reflected vision than in direct vision. One example for all of these [cases] may be applied, and the same should be understood [to apply] to the rest. 
◉ In visu directo, cum fuerit corpus visum remotum ab axibus visualibus, accidit ipsum videri duo; idem evenit in speculis re visa ab axibus elongata. 
◉ In direct vision, when the visible object lies far outside the visual axes, it may appear double; the same thing happens in mirrors when the visible object lies far outside the visual axes. 
◉ In speculis ab aliqua longitudine videbitur corpus minus quam sit, quod forsan directe a tanta longitudine videretur minus quam esset in veritate, sed non adeo minus. Et hoc minoritatis additamentum in speculis provenit propter minus longitudinis temperamentum. 
◉ In mirrors, the object will appear smaller than it should at a given distance, whereas at such a distance it may look smaller than it should in direct vision, but not to such a great extent.⁑ And this increased diminution [which happens] in mirrors is due to the decrease in the [range of] threshold conditions [for the perception] of distance. 
◉ In figura non numquam accidit error in speculis per causas per quas in directo, sed maior et frequentior propter situm. 
◉ In [the perception of] shape error sometimes arises in mirrors for the same reasons it does in direct vision, but [it does so] more significantly and more frequently according to spatial disposition. 
◉ Si aliquid ab aliqua longitudine opponatur speculo, et eius capita non percipiantur a visu, ut funis vel aliquid tale, videbitur forsan continuum speculo. Idem accidit in visu directo. Si opponatur funis aliquis foramini et non videantur capita funis, non apparebit distantia inter funem et foramen, licet magna sit, et est propter situm. Si autem alterum capitum videatur, aliud vero non, videbitur fortassis illud caput continuum. Et in singulis ubi directe accidit similiter in reflexione. 
◉ If a rope or something like it faces a mirror at a given distance, and if its ends cannot be perceived by the visual faculty, it may appear to lie on the very surface of the mirror. The same thing happens in direct vision. If some rope is placed facing a window and the ends of the rope cannot be seen, the separation between rope and window will not be apparent, even if it is significant, and [this] is due to spatial disposition.⁑ Moreover, if one of the ends is visible but the other is not, that end may appear to lie in the plane [of the window]. In each case, where [error] occurs in direct vision, it occurs likewise in reflected vision. 
◉Pars quarta 
◉CHAPTER 4 
In speculis spericis [exterioribus] 

◉ Universitas errorum in speculis planis accidentium evenit similiter in spericis exterioribus, et preter hoc, in spericis speculis res visa videtur minor quam sit. Et generaliter in hiis speculis nichil ex re visa comprehenditur in veritate preter ordinationem partium, que talis apparet in speculo qualis est in corpore viso. 
◉ The entire range of errors that occur in plane mirrors also occurs in convex spherical [mirrors], and besides this, a visible object looks smaller than it should in [convex] spherical mirrors. Overall, in these [kinds of] mirrors nothing about the visible object is perceived as it actually is except the arrangement of its parts, which appears in the mirror as it actually exists in the visible object. 
◉ [PROPOSITIO 2] Quod autem semper videatur res minor in hoc speculo quam ipsa sit probatur. 
◉ [PROPOSITION 2] That an object should always appear smaller than it is in this [sort of] mirror is demonstrated [as follows]. 
◉ Sit AB [FIGURE 6.4.2, p. 304] linea visa, ZP speculum, D centrum circuli, E centrum visus. A reflectatur ad E a puncto H, B a puncto N. Linea AB producta aut transibit per centrum speculi, aut non. 
◉ Let AB [in figure 6.4.2, p. 97] represent a visible line [on some object, let] ZP be the mirror, D the center of the [great] circle [produced by the plane of reflection on the mirror], and E the center of sight. Let [the form of point] A be reflected to [point] E from point H, and [let the form of point] B [be reflected to E] from point N. When it is extended, line AB will pass through the center of the mirror, or [it will] not. 
◉ Transeat. Et ducatur a puncto N linea contingens circulum, que sit NL; a puncto H contingens HM. Et ducantur linee reflexionis BN, EN, AH, EH, et producantur linee EH, EN donec cadant in perpendicularem, que est AD, et puncta casus sint T, Q. Palam quoniam T est locus ymaginis A; Q est locus ymaginis B. Dico quoniam AB maior est QT. 
◉ Let it pass through. From point N draw line NL tangent to the circle, and from point H [draw] tangent HM. Draw the line[couple]s of reflection BN and EN, and AH and EH, extend lines EH and EN until they fall on normal AD, and let T and Q be the points where they fall. It is evident that T is the imagelocation for A, and Q is the imagelocation for B. I say that AB > QT. 
◉ Patet ex superioribus quoniam proportio AD ad DT sicut AM ad MT. Similiter, proportio BD ad DQ sicut proportio BL ad LQ. Sed AD maior BD, et DT minor DQ. Erit igitur maior proportio AM ad MT quam BL ad LQ. 
◉ It is evident from previous discussions [in book 5, prop. 7] that AD:DT = AM:MT. Likewise, BD:DQ = BL:LQ.⁑ But AD > BD, and DT < DQ [so AD:DT > BD:DQ and thus > BL:LQ]. Hence, AM:MT [which = AD:DT] > BL:LQ. 
◉ Secetur AM in puncto F ut proportio FM ad MT sit sicut BL ad LQ. Erit igitur minor BM ad MT quam BL ad LQ. Secetur MT in puncto K ut proportio BM ad MK sit sicut BL ad LQ. K necessario cadet inter M et Q, quoniam LQ minor MQ, et BL maior BM. Cum igitur FM ad MT sicut BL ad LQ et sicut BM ad MK, erit proportio FB ad KT sicut BL ad LQ. Sed BL maior LQ. Igitur FB maior KT, quare AB maior QT, quod est propositum. 
◉ Cut AM at point F so that FM:MT = BL:LQ. Therefore, BM:MT < BL:LQ. Cut MT at point K so that BM:MK = BL:LQ. K will necessarily fall between M and Q because LQ < MQ, and BL > BM. Accordingly, since FM:MT = BL:LQ, as well as BM:MK, FB:KT = BL:LQ.⁑ But BL > LQ [because we know by previous conclusions that BD:DQ = BL:LQ, and BD > DQ, so BL > LQ]. Hence, FB > KT, so [visible line] AB > [its image] QT [since AB > FB, and QT < KT], which is what was set out [to be proven]. 
◉ Si vero linea AB producta non perveniat ad centrum, ducatur a puncto A [FIGURE 6.4.2a, p. 304] linea ad centrum, que sit AG, et sit G centrum, et a puncto B ducatur linea BG. Locus ymaginis A sit punctus D, locus ymaginis B sit E, et ducatur linea ED, que quidem est ymago linee AB. Dico quoniam AB maior est ED, quoniam ED aut est equidistans AB aut non. 
◉ But if line AB, when it is extended, does not reach the center [of the circle], then from point A [in figures 6.4.2a and 6.4.2b, p. 98] draw line AG to the center, let G be the center, and from point B draw line BG. Let point D be the imagelocation for [point] A, let [point] E be the imagelocation for [point] B, and draw line ED, which is the image of line AB. I say that [object] AB > [image] ED because ED is either parallel to AB or not. 
◉ Si fuerit equidistans, planum quoniam est minor. Si non fuerit equidistans, producatur usquoque concurrat cum ea. Sit concursus Z, et a puncto E ducatur equidistans AB, que sit EH. Angulus EDH aut est acutus, aut rectus, vel maior. 
◉ If it is parallel [as in figure 6.4.2a], it is clear that it is smaller [i.e., ED < AB because triangles EDG and BAG are similar, so BA:ED = BG:EG, and BG > EG]. If it is not parallel [as in figure 6.4.2b], extend [ED] until it meets AB. Let Z be the [point of] intersection, and from point E draw EH parallel to AB. Angle EDH is acute, right, or greater [than a right angle]. 
◉ Si rectus vel maior, erit latus EH maius ED. Sed EH minus AB, et ita propositum. 
◉ If it is right or greater [than a right angle], side EH > [side] ED. But [by previous conclusions] EH < AB, and so [we have demonstrated] what was set out [to be proven].⁑ 
◉ Si fuerit acutus, poterit accidere quod forma sit maior ipsa re cuius est forma, quam licet excedat raro accidet. Et cum acciderit, forsan comprehendetur forma a longitudine tali quod minor videbitur quam sit, quoniam ipsum corpus ab hac longitudine forsan videtur minus. 
◉ If it is acute, it could happen that the form [i.e., ED] is larger than the object [AB] whose form it is, which, although it may be larger, will happen rarely. And when it does happen, the form may be perceived from such a distance that it will appear smaller than it should because the object itself may appear smaller [than it should] at that distance [in direct vision].⁑ 
◉ [PROPOSITIO 3] Quod autem forma in hiis speculis aliquando videatur maior re visa, scilicet cum maior fuerit, et comprehendatur a tali longitudine a qua certa eius quantitas possit discerni declarabitur. 
◉ [PROPOSITION 3] It will now be demonstrated that in these [kinds of] mirrors a form may sometimes appear larger than the visible object, i.e., when it [actually] is larger, and that it may be perceived [as larger] from such a distance that its size can be discerned with proper certitude. 
◉ Sit A [FIGURE 6.4.3, p. 305] centrum speculi, et superficies sumatur reflexionis que secabit speculum super circulum. Sit circulus ille EDB, ED dyameter illius circuli, et producatur dyameter ED usque ad Z ut multiplicatio EZ in ZD non sit maior quadrato AD, quod planum, cum sit possibile dyametro ED talem addi lineam ut ductus totalis in partem additam sit equalis quadrato AD. Et dividatur linea ZD in partes equales in puncto H. Erit igitur AH medietas EZ. Ductus ergo AD in HD non erit maior quarta parte quadrati AD, et quoniam ductus AH in HD maior est quadrato HD, sit ductus AH in HT equale quadrato HD. 
◉ Let A [in figure 6.4.3, p. 99] be the center of the mirror, and take a plane of reflection that will cut the mirror along a [great] circle. Let that circle be EDB, let ED be the diameter of that circle, and extend diameter ED to Z so that rectangle EZ,ZD is not greater than AD2, which is clear[ly possible], since it is possible for a line to be added to diameter ED such that the rectangle formed by the whole and the added part equals AD2 [by Euclid, III.36].⁑ Bisect line ZD at point H. Hence, AH will be half of EZ. Accordingly, [since AD < AH, which = half EZ, while DH = half DZ] rectangle AD,HD will not be greater than onefourth AD2 [because it is no greater than onefourth rectangle EZ,ZD, which is no greater than AD2], and since AH,HD > HD2, let AH,HT = HD2.⁑ 
◉ Fiat circulus secundum quantitatem AH, et a puncto H ducatur corda equalis medietati linee HD, que sit HQ. Et producantur linee QA, QT, et supra punctum Q fiat angulus equalis angulo QAH, qui sit HQN. Cum ergo hiis duobus triangulis hii duo anguli sint equales, et unus communis, scilicet QHA, erit tertius tertio equalis, scilicet AQH angulo HNQ. Et erunt trianguli similes, et erit proportio AH ad HQ sicut HQ ad HN. Igitur quod fit ex ductu AH in HN equale quadrato HQ. 
◉ Produce a circle according to length AH [as radius], and from point H draw chord HQ equal to onehalf line HD. Draw lines QA and QT, and at point Q form angle HQN equal to angle QAH. Accordingly, since these two angles in these two triangles [HQN and QAH] are equal, and since one [angle], i.e., QHA, is common, the third [angle] = the third [angle], i.e., [angle] AQH = angle HNQ. And [so] the triangles will be similar [by Euclid, VI.4], and [according to proportional sides] AH:HQ = HQ:HN. Therefore, AH,HN = HQ2 [by Euclid, VI.17]. 
◉ Sed quadratum HQ est quarta pars quadrati HD, cum HQ sit medietas HD. Igitur multiplicatio AH in HN equalis est quarte parti multiplicationis AH in HT, quare HN est quarta pars HT. Igitur N cadit inter H et T. Restat ut ductus HT in TN sit tres quarte quadrati HT. 
◉ But HQ2 is onefourth HD2, since HQ is onehalf HD [by construction]. Therefore, AH,HN = onefourth AH,HT [which = HD2, by construction], so HN is onefourth HT. Accordingly, N lies between H and T. It follows that HT,TN is threefourths HT2. 
◉ Verum angulus QHD acutus, et equalis angulo HQA, quia respiciunt equalia latera in maiori triangulo. Igitur angulus QHN equalis angulo HNQ, et ita HQ equalis QN. 
◉ Angle QHD is acute, however, and [it is] equal to angle HQA, since they are subtended by equal sides in the larger triangle [i.e., QA and HA, which are radii of the larger circle]. Therefore, angle QHN [in triangle AQH] = angle HNQ [in triangle HQN, which is similar to triangle AQH, by previous conclusions], and so HQ = QN. 
◉ Et angulus HNQ acutus, quare angulus QNT obtusus. Quadratum igitur TQ superat quadratum QN et quadratum TN ductu linee TN in NH, quoniam, ut dicit Euclides, quadratum lateris oppositi obtuso superat quadrata duorum laterum quantum est quod fit ex ductu unius lateris bis in partem ei adiunctam procedentem usque ad locum casus perpendicularis a capite alterius lateris ducte. Et si a puncto Q ducatur perpendicularis super lineam HT, cadet in puncto medio linee HN, et ductus TN in medietatem HN bis equipollet ductui TN in HN. 
◉ Also, angle HNQ is acute, so [adjacent] angle QNT is obtuse. Hence, TQ2 exceeds QN2 + TN2 by TN,NH because, as Euclid claims [in II.12], the square on the opposite side of an obtuse [angle] exceeds the squares on the [other] two sides by twice the rectangle formed by one of the sides and the adjoining segment that extends to where the perpendicular is dropped [to it] from the endpoint of the other side. And if a perpendicular is dropped from point Q to line HT, it will fall at the midpoint of line HN [because triangle QNT is isosceles], and twice the rectangle formed by TN and onehalf HN equals TN,HN. 
◉ Igitur quadratum TQ superat quadratum QN et TN ductu TN in NH. Sed ductus HN in NT cum quadrato NT equalis est ductui HT in TN. Igitur ductus HT in TN est excessus quadrati TQ supra quadratum HQ. 
◉ Therefore, TQ2 exceeds QN2 + TN2 by TN,NH [i.e., TQ2 – QN2 – TN2 = NH,NT, so NH,NT + TN2 = TQ2 – QN2]. But HN,NT + NT2 = HT,TN [by Euclid, II.3]. Therefore HT,TN = TQ2 – HQ2 [because HQ2 = QN2, since HQ = QN in isosceles triangle QHN]. 
◉ Amplius, sit proportio AI ad AH sicut QT ad QH [FIGURE 6.4.3a, p. 305]. Erit quadratum ad quadratum sicut quadratum ad quadratum, et erit proportio excessus quadrati AI super quadratum AH ad quadratum AH sicut ductus HT in TN ad quadratum QH. Et quoniam quadratum QH quater sumptum efficit quadratum HD, et ductus HT in TN quater sumptus efficit triplum quadrati HT, erit ductus HT in TN ad quadratum QH sicut tripli quadrati HT ad quadratum HD. 
◉ Now let AI:AH = QT:QH [by Euclid VI.12]. The square [of AI] to the square [of AH] will be as the square [of QT] to the square [of QH—i.e., AI2:AH2 = QT2:QH2], and (AI2 – AH2):AH2 = HT,TN [which = QT2 – QH2]:QH2 [by Euclid, V.17]. And since 4QH2 = HD2 [by previous conclusions], while 4HT,TN = 3HT2 [by previous conclusions], HT,TN:QH2 = 3HT2 [which = 4HT,TN]:HD2 [which = 4QH2]. 
◉ Sit autem HC tripla ad HT. Erit ductus CH in HA triplus ad quadratum HD, sed quoniam proportio AH ad HD sicut HD ad HT, erit HT ad HA sicut quadratum HT ad quadratum HD. Verum proportio CH ad HA sicut ductus CH in HT ad ductum HA in HT, et ita CH ad HA sicut tripli quadrati HT ad quadratum HD. Sed hec erat proportio excessus quadrati AI super quadratum AH ad quadratum AH. Igitur CH ad HA sicut excessus quadrati AI super quadratum AH ad quadratum AH. Igitur coniunctim proportio CA ad AH sicut quadrati AI ad quadratum HA, excessus enim quadrati AI super quadratum HA cum quadrato HA efficit quadratum AI. 
◉ Moreover, let HC = 3HT. [Therefore,] CH,HA = 3HD2 [since HT,HA = HD2, by construction], but since AH:HD = HD:HT [because AH,HT = HD2, by construction, so HD is the mean proportional between AH and HT], HT:HA = HT2:HD2.⁑ But CH:HA = CH,HT:HA,HT, and so CH:HA = 3HT2:HD2 [because CH:HA = 3HT:HA]. But this [i.e., 3HT2:HD2] was as (AI2 – AH2):AH2.⁑ Therefore, CH:HA = (AI2 – AH2):AH2. Therefore, CA [which = CH + HA]:AH = AI2:HA2, for (AI2 – HA2) + HA2 = AI2. 
◉ Igitur IA erit media in proportione inter CA et HA, cuius rei conversam paulo ante tetigimus. Igitur proportio CA ad IA sicut IA ad HA, et eadem erit proportio residui ad residuum, id est CI ad IH, et cum IA maior HA, erit CI maior IH. 
◉ Hence, IA will be the mean proportional between CA and HA, whose converse we touched upon a bit earlier.⁑ Accordingly, CA:IA = IA:HA, and the remainder will be in the same proportion to the remainder, i.e., CI:IH [= CA:IA = IA:HA], and since IA > HA, CI > IH. 
◉ Amplius, ductus AH in HD minor quarta parte quadrati AD. Igitur HD est minor quarta parte linee AD. Igitur est minor quinta parte AH. Cum ergo AH sit maior quam quintupla ad HD, et ductus eius in HT efficiat quadratum HD, erit HT minor quinta parte HD, et ita HT erit minor vicesima quinta parte HA. Sed proportio CI ad IH sicut IA ad HA, ut dictum est. Igitur, coniunctim erit CH ad IH sicut IA cum AH ad AH. Igitur tertia primi ad secundum sicut tertia tertii ad quartum. 
◉ Furthermore, AH,HD < onefourth AD2 [by previous conclusions]. Therefore, [line] HD < onefourth line AD [because AH > AD]. So it is less than onefifth AH [because AH = AD + HD]. Therefore, since AH > 5HD, and since AH,HT = HD2 [by construction], HT < onefifth HD, and so HT < one twentyfifth HA. But, as was [just] claimed, CI:IH = IA:HA. Thus, CH [which = CI + IH]:IH = (IA + AH):AH. Hence, onethird the first [term is] to the second as onethird the third [term is] to the fourth [i.e. onethird CH:IH = onethird (IA + AH):AH]. 
◉ Sed HT est tertia pars linee CH. Igitur TH ad IH sicut tertia pars linee IA cum AH ad lineam AH. Igitur TH ad IH sicut due tertie linee AH cum tertia linee IH ad lineam AH. Sed quoniam CI maior IH, erit IH minor medietate CH, et erit tertia IH minor sexta parte CH, et ita tertia IH erit minor medietate TH. Igitur due tertie AH cum minori parte medietate HT se habebunt ad AH sicut TH ad IH. Igitur IH ad HT sicut AH ad duas sui tertias cum minori parte medietate HT. 
◉ But [line] HT is onethird line CH [by construction]. Therefore, TH:IH = onethird (line IA + AH):line AH. Accordingly, TH:IH = (twothirds line AH + onethird line IH):line AH.⁑ However, since CI > IH [from previous conclusions], IH < onehalf CH, and onethird IH < onesixth CH, and so onethird IH < onehalf TH [which = onethird CH]. Therefore, (twothirds AH + less than onehalf HT):AH = TH:IH. So, conversely, IH:HT = AH:(twothirds AH + less than half HT). 
◉ Sed HT minor vicesima quinta AH, et eius medietas minor medietate vicesime quinte. Sed linea AH in viginti quinque partes divisa; due eius tertie cum medietate vicesime quinte non efficiunt octodecim eius partes. Igitur proportio IH ad HT maior quam sit proportio viginiti quinque ad octodecim. 
◉ But HT < one twentyfifth AH [by previous conclusions], and its half < half of one twentyfifth. But line AH is divided into twentyfive parts, [so] twothirds [of those 25 parts, i.e., < 17] + half of one twentyfifth [part] does not add up to 18 parts. Therefore, IH:HT > 25:18. 
◉ Item, cum HT sit minor vicesima quinta AH, erit AT maior viginti quattuor vicesime quinte AH. Sed linea IH minor medietate CH, et ita minor HT cum medietate HT, et ita minor una et dimidia viginti quinque partium AH, et ita IA minor viginti sex et dimidie partis sumptis partibus secundum divisionem HA in viginti quinque. Ergo proportio IA ad AT sicut minoris viginiti sex et dimidii ad maius viginti quattuor. Igitur proportio IA ad AT minor quam viginti sex et dimidii ad viginti quattuor. Sed IH ad HT maior quam viginiti quinque ad octodecim. Igitur IH ad HT maior quam IA ad AT. 
◉ Furthermore, since HT < one twentyfifth AH, AT > twentyfour twentyfifths AH. But line IH < onehalf CH, so [IH] < (HT + onehalf HT), so [IH] < one and onehalf of [one of] the twentyfive parts comprising AH, and so IA < twentysixandonehalf of the given 25 parts into which HA is divided. Therefore, IA:AT = (less than twentysixandahalf):more than 24. Thus, IA:AT < twentysixandahalf:24. But IH:HT > 25:18. Therefore, IH:HT > IA:AT. 
◉ Sit proportio IM ad MT sicut IA ad AT. Cadet quidem M inter I, H. Item, maior erit proportio IM ad MH quam IA ad AT, et ita maior quam IA ad AH. Sit igitur proportio IL ad LH sicut IA ad AH. Cadet quidem L inter M et I. 
◉ [By Euclid VI.12] let IM:MT = IA:AT [in figure 6.4.3a, p. 100]. M will therefore fall between I and H. Moreover, IM:MH > IA:AT, and so it is greater than IA:AH. So let IL:LH = IA:AH [by Euclid VI.12]. L will of course fall between M and I. 
◉ Amplius, a punctis L, M ducantur contingentes LB, MG, et ducantur linee IB, HB, IG, TG, AB, AG, que ultime producantur usque ad exteriorem circulum. 
◉ Now from points L and M draw tangents LB and MG, and draw lines IB, HB, IG, TG, AB, and AG, and extend the last [two] to the outer circle [to intersect it at points Z2 and Z1, respectively].⁑ 
◉ Et habebis ex quinta quinti libri quod angulus IBZ sit equalis angulo HBA, cum enim sit proportio IL ad LH sicut IA ad AH, erit H locus ymaginis in reflexione a puncto B. Et si dicatur contrarium ut sumatur alius locus ymaginis, improbabis per impossibile, sumpta impossibilitate a proportione quam verum est esse IA ad lineam a puncto A ad locum ymaginis sicut IL ad lineam a puncto L ad locum ymaginis. 
◉ From the fifth [proposition] of the fifth book you will [then] have that angle IBZ2 = angle HBA, for, since IL:HL = IA:AH [by construction], H will be the imagelocation [of objectpoint I] in the case of reflection from point B [when HB, extended beyond B to H’, forms the line of reflection].⁑ And if the contrary is claimed, so that some other imagelocation is chosen, you will disprove it by a reductio ad absurdum, given that it is impossible for the ratio of IA to the line from point A to the imagelocation not to be as [the ratio of] IL to the line from point L to the imagelocation. 
◉ Cum igitur H sit locus ymaginis, et LB sit contingens super AB, producta HB faciet angulum reflexionis equalem sibi collaterali, et quoniam LB perpendicularis super ABZ, restabit angulus IBL equalis angulo LBH. Eodem modo erit angulus IGZ equalis angulo TGA, et cum MG sit perpendicularis, erit angulus IGM equalis angulo MGT. 
◉ Therefore, since H is the imagelocation, and since LB is tangent [to the mirror] on AB, then when it is extended [to H’], HB will form an angle of reflection [H’BZ2] equal to its vertical [angle HBA], and because LB is perpendicular to ABZ2, it will follow that angle IBL = angle LBH. By the same token, angle IGZ1 = angle TGA, and since MG is perpendicular [to AG], angle IGM = angle MGT. 
◉ Amplius, producatur a puncto H ad lineam AB linea equidistans IB, que sit HP, et a puncto T equidistans IG, que sit TR. Erit angulus IBZ equalis angulo HPB. Sed angulus IBZ equalis, ut dictum est, angulo HBA, et ita duo anguli HBA, HPB sunt equales, quare duo latera HB, HP sunt equalia. Similiter, TR equalis TG. Verum angulus HPB acutus, cum sit equalis angulo reflexionis; erit angulus HPA obtusus, et erit HA maior HP, et ita maior HB. Similiter, erit TA maior TG. 
◉ Now draw line HP from point H to line AB parallel to [line] IB, and from point T [draw] TR parallel to IG. Angle IBZ2 = angle HPB. But, as was claimed [earlier], angle IBZ2 = angle HBA, and so the two angles HBA and HPB are equal, so the two sides HB and HP [of triangle HBP] are equal. Likewise [in triangle TGR, side] TR = [side] TG. But angle HPB is acute, since it is equal to the angle of reflection [IBZ2, so adjacent] angle HPA will be obtuse, and HA > HP [by Euclid, I.19], so it will be greater than HB [since HB = HP]. So too, TA > TG. 
◉ Amplius, quoniam HP equidistans IB, erit proportio IA ad AH sicut AB ad AP; erit similiter proportio IA ad AT sicut GA ad AR, et erit proportio AH ad AI sicut AP ad AB. Sed IA ad AT sicut AB ad AR, cum AB sit equalis AG. Igitur a primo erit proportio AH ad AT sicut AP ad AR. 
◉ Moreover, since HP is parallel to IB, then IA:AH = AB:AP [by Euclid, VI.2]; likewise, IA:AT = GA:AR, and [conversely] AH:AI = AP:AB. But IA:AT = AB:AR, since AB = AG. Therefore, from the first, AH:AT = AP:AR [by Euclid, V.22].⁑ 
◉ Verum, cum angulus HPA obtusus, quadratum HA excedet quadratum HP et quadratum AP cum multiplicatione AP in lineam ductam a puncto P usque ad locum perpendicularis ducte a puncto H bis. Sed perpendicularis ducta a puncto H cadet in puncto medio linee PB, cum HB, HP sint equales, et ita quadratum HA excedet quadratum HP et quadratum AP in multiplicatione AP in PB. Et ita quadratum AH excedit quadratum HP in multiplicatione AB in AP, quoniam ductus AP in PB cum quadrato AP valet ductum AB in AP. Similiter, quadratum AT excedit quadratum TR in ductu AG in AR, sive AB in AR, quod idem est. 
◉ But since angle HPA [in triangle HPA] is obtuse, HA2 will exceed (HP2 + AP2) by twice the rectangle formed by AP and the line [segment] extended from point P to the perpendicular dropped [to AP] from point H [by Euclid, II.12]. But the perpendicular dropped from point H [to AP] will fall to the midpoint of line PB, since HB and HP are equal [as established earlier], and so HA2 will exceed (HP2 + AP2) by AP,PB [i.e., HA2 – HP2 – AP2 = AP,PB, so HA2 – HP2 = AP2 + AP,PB]. Accordingly, AH2 exceeds HP2 by AB,AP [i.e., AH2 – HP2 = AB,AP] because (AP,PB + AP2) = AB,AP. Likewise, AT2 exceeds TR2 by AG,AR, or AB,AR, which is identical [since AG = AB, both being radii of the circle]. 
◉ Ducatur ergo linea AB in duas lineas AP, AR, et provenient duo excessus. Igitur proportio excessus ad excessum sicut AP ad AR, quare proportio excessus quadrati AH super quadratum HP ad excessum quadrati AT super quadratum TR sicut AH ad AT. Et cum HP equalis HB, et TR, TG, erit proportio excessus quadrati AH super quadratum HB ad excessum quadrati AT super quadratum TG sicut AH ad AT. 
◉ Accordingly, combine line AB with the two lines AP and AR [to form rectangles AB,AP and AB,AR], and the two remainders will be formed. Hence, remainder [AB,AP] is to remainder [AB,AR] as AP:AR, so (AH2 – HP2 [which = AB,AP]):(AT2 – TR2 [which = AB,AR]) = AH:AT [since AB,AP:AB,AR = AP:AR = AH:AT, by previous conclusions]. And since HP = HB, and TR = TG, (AH2 – HB2):(AT2 – TG2) = AH,AT. 
◉ Sed multiplicatio AH in HU equalis est quadrato HB. Igitur multiplicatio AH in AU erit excessus quadrati AH super quadratum HB. Igitur proportio AH ad AT sicut multiplicationis AH in AU ad excessum quadrati AT super quadratum TG. Et si due linee AH, AT ducantur in AU, erit proportio AH ad AT sicut multiplicationi AH in AU ad multiplicationem AT in AU. Igitur multiplicatio AT in AU est excessus quadrati AT super quadratum TG. Igitur erit multiplicatio AH in HU equalis quadrato HB, et multiplicatio AT in TU equalis quadrato TG. 
◉ But [let us cut line AH at point U such that] AH,HU = HB2. [However, AH2 = AH,AU + AH,HU (by Euclid, II.2), so AH2 – AH,HU (which = HB2) = AH,AU]. Therefore, AH,AU = AH2 – HB2. So AH:AT = AH,AU:(AT2 – TG2) [since AH:AT = (AH2 – HB2):(AT2 – TG2), by previous conclusions, and (AH2 – HB2) = AH,AU]. And if the two lines AH and AT are combined with AU [to form rectangles, then] AH:AT = AH,AU:AT,AU. Hence, AT,AU = AT2 – TG2. So AH,HU = HB2 [by construction], and AT,TU = TG2 [since AT2 = AT,AU + AT,TU, and, consequently, AT,AU = AT2 – AT,TU. But because AT,AU = AT2 – TG2, then AT,TU = TG2]. 
◉ Amplius, arcus BG dividatur per equalia in puncto O, et ducantur tres perpendiculares super lineam HA, scilicet BF, OY, GK, et a puncto G ducatur linea equidistans HA, que sit GS, et a puncto B ducatur perpendicularis super AG, que sit BX. Hoc quidem BX, si produceretur usque ad circulum, divideret linea AG ipsam per equalia, et arcum cuius esset corda. Et ita secaretur alius arcus equalis arcui BG, quoniam alium arcum respiceret angulus GBX, et ita angulus GBX est medietas anguli supra centrum respicientis eundem arcum, secundum Euclidem. Igitur angulus GBX est medietas anguli BAG quem dividit linea OA per equalia. Igitur angulus GBX est equalis angulo OAG. Duo autem anguli BSG, BXG recti. 
◉ Now bisect arc BG at point O [see inset to figure 6.4.3a, p. 100], drop the three perpendiculars BF, OY, and GK to line HA, draw line GS from point G parallel to HA, and drop perpendicular BX from point B to AG. If BX were extended to the circle [DGB], line AG would bisect it, as well as the arc whose chord it would form [when extended]. Accordingly, it[s other half after extension] would cut off another arc equal to arc BG, since its other arc would subtend angle GBX, and so angle GBX is half the angle [GAB] at the center [of the mirror] subtended by that same arc, according to Euclid [III.20]. Hence, angle GBX is onehalf angle BAG, which line OA bisects. Therefore, angle GBX = angle OAG. Moreover, the two angles BSG and BXG are right [by construction]. 
◉ Si intelligatur circulus super BG transiens per S, transibit per X, et fiet arcus SX super quem cadent duo anguli XBS, XGS. Igitur hii duo anguli sunt equales. Sed angulus GAY equalis angulo XGS propter equidistantiam linearum, et ita angulus GAY equalis angulo XBS. Et ut dictum est, angulus GBX equalis angulo OAG. Erit angulus OAY equalis angulo GBS, et erit triangulus OAY similis triangulo GBS. Igitur proportio GB ad BS sicut OA ad AY. 
◉ If a circle is imagined on BG [as diameter, and if it] passes through S, it will pass through X [by Euclid, III.31], and arc SX will be formed such that the two angles XBS and XGS [i.e., AGS, since X lies on AG] will fall upon it. These two angles will therefore be equal [by Euclid, III.27]. But angle GAY = [alternate] angle XGS because of the parallelism of lines [AY and GS], so angle GAY = angle XBS. And, as has [already] been claimed, angle GBX = angle OAG. Angle OAY = angle GBS, and [so] triangle OAY will be similar to triangle GBS.⁑ Hence, GB:BS = OA:AY. 
◉ Amplius, cum angulus AHB sit acutus, quadratum AB minuit ex quadratis AH, HB quantum est illud quod fit ex ductu AH in HF bis, secundum quod dicit Euclides. Igitur quadratum AH cum quadrato HB superat quadratum DA, que est equalis AB, in ductu AH in HF bis, et ita in ductu AH in HD bis et AH in DF bis. Sed multiplicatio AH in HD bis cum quadrato AD est equalis quadrato AH cum quadrato HD. Et ita ablato communi quadrato AB [cum multiplicatione AH in HD bis], restabit quadratum HD cum ductu AH in FD bis equalis quadrato HB. 
◉ Now since angle AHB [in triangle AHB] is acute, AB2 is less than AH2 + HB2 by 2AH,HF [i.e., AH2 + HB2 – AB2 = 2AH,HF], according to what Euclid claims [in II.13]. Therefore, [since] DA = AB [because they are radii of circle DGB], AH2 + HB2 is greater than DA2 by 2AH,HF [i.e., AH2 + HB2 – DA2 (i.e., AB2) = 2AH,HF], and thus by 2AH,HD + 2 AH,DF [i.e., AH2 + HB2 – DA2 = 2AH,HD + 2 AH,DF, or AH2 + HB2 = DA2 + 2AH,HD + 2AH,DF]. But 2AH,HD + AD2 = AH2 + HD2 [by Euclid, II.7]. So, if the common [term] (AB2 [+ 2AH,HD]) is subtracted, it will follow that HD2 + 2AH,FD = HB2.⁑ 
◉ Sed multiplicatio AH in HT equalis est quadrato HD, et multiplicatio AH in HU equalis quadrato HB. Erit ergo multiplicatio AH in HU equalis multiplicationi AH in HT et multiplicationi AH in DF bis. Subtracto ductu AH in HT, quem communem ponimus utrique multiplicationi, restabit multiplicatio AH in TU equalis multiplicationi AH in DF bis. Igitur TU est dupla DF. 
◉ But AH,HT = HD2 [by construction], and AH,HU = HB2 [by construction]. Hence, AH,HU = AH,HT + 2AH,DF. Having subtracted AH,HT, which we designate as common to both rectangles, it will follow that AH,TU = 2AH,DF. So TU = 2DF. 
◉ Amplius, cum angulus ATG sit acutus, secundum predictum modum erit quadratum AT cum quadrato TG equale quadrato AD cum ductu AT in TK bis, et ita cum ductu AT in TD bis, et in DK bis. Et probatur modo predicto quod quadratum TG equale est quadrato TD cum ductu AT in DK bis. Sed ductus AT in TU equalis quadrato TG, et ita equalis quadrato TD cum ductu AT in DK bis. 
◉ Moreover, since angle ATG is acute, then according to previous reasoning [based on Euclid, II.13], AT2 + TG2 = AD2 [which = AG2] + 2AT,TK, and so [AT2 + TG2 = AD2] + 2AT,TD + 2AT,DK. And it is proven by previous reasoning that TG2 = TD2 + 2AT,DK.⁑ But AT,TU = TG2 [by previous conclusions], and so AT,TU = TD2 + 2AT,DK. 
◉ Sit ductus AT in TE equalis quadrato TD. Restat ergo ut ductus AT in EU sit equalis ductui AT in DK bis, per ablationem communis, quod est ductus AT in TE. Igitur EU est dupla DK. Sed iam dictum est quod TU est dupla DF. Restat ergo TE duplum FK. [4.40] Amplius, proportio AH ad HT sicut AH ad HD duplicata, HD enim media est in proportione inter illas, cum eius quadratum sit equale ductui AH in HT. Et similiter proportio AT ad TE sicut AT ad TD duplicata. Sed maior est proportio AT ad TE quam AH ad HD. Igitur maior est proportio AT ad TE quam AH ad HT, et cum AH maius AT, erit HT maior TE. Sed TE dupla FK. 
◉ Let [point E’ be chosen on line AT such that] AT,TE’ = TD2. It follows, then, that AT,E’U [which = AT,TU – AT,TE’] = 2AT,DK [which = TD2 + 2AT,DK – AT,TE’ (which = TD2)], if the common term AT,TE’ is subtracted. Thus, E’U = 2DK. But it has already been claimed that TU = 2DF. It follows, then, that TE’ [which = TU – E’U] = 2FK [which = 2DF – 2DK]. 
◉ Item, ut dictum est, proportio BG ad GS sicut OA ad OY. Erit proportio BG ad OA sicut GS ad OY. Sed OA equalis BA, et GS equalis FK. propter equidistantiam. Erit proportio BG ad BA sicut FK ad OY. 
◉ Moreover, as has been claimed [earlier], BG:GS = OA:OY. [So] BG:OA = GS:OY. But OA = BA [since they are radii of circle DGB], and GS = FK, according to the parallelism [of lines GS and FK and lines GK and FS]. Hence, BG:BA = FK:OY. 
◉ Amplius, IH minor medietate CH, et CH tripla HT. Erit IH minor HT et medio ipsius. Sed HT minor quinta HD. Igitur IH minor TD multo, quare IH multo minor ND, quare MI minor ND. Et palam per hoc quod I cadet inter H et Z. 
◉ In addition, IH < onehalf CH [by previous conclusions], and CH = 3HT [by construction]. [So] IH < oneandonehalf HT. But HT < onefifth HD [by previous conclusions]. Accordingly, IH is smaller yet than TD, so IH is even smaller than ND, and so MI < ND [since MI < IH]. From this it is evident that I will lie between H and Z. 
◉ Amplius, quod fit ex ductu EZ in ZD non est maius quadrato AD; igitur quod fit ex ductu EM in MD est minus quadrato AD. Sed quoniam MG est contingens, quod fit ex ductu EM in MD est equale quadrato MG, secundum quod dicit Euclides. Igitur MG est minor AD; igitur MG est minor AG. 
◉ Furthermore, EZ,ZD is not greater than AD2 [by construction, with EH as the diameter of circle QHZ], so EM,MD < AD2. But since MG is tangent [to circle DGB], EM,MD = MG2, according to what Euclid claims [in III.36]. Thus, MG < AD, so MG < AG. 
◉ Amplius, duo trianguli AGM, MGK habent unum angulum communem, et uterque eorum habet angulum rectum. Igitur sunt similes, quare proportio MK ad KG sicut MG ad GA, et ita MK minor est KG. Et cum OY sit maior GK, erit HD minor OY. 
◉ Also, the two triangles AGM and MGK have a common angle [i.e., AMG], and both of them have a right angle [i.e., MGA and MKG]. Hence, they are similar [by Euclid, VI.4], so MK:KG = MG:GA, and so MK < KG [since MG < GA, by previous conclusions]. And since OY > GK, HD < OY [since HD < GK < OY]. 
◉ Amplius, AH ad HD sicut HD ad HT; erit ergo sicut medietas HD ad medietatem HT. Et ita AH ad HD sicut QH ad medietatem HT, cum QH sit medietas HD, et ita AH ad QH sicut HD ad medietatem HT, et ita QH ad AH sicut medietas HT ad HD. Sed medietas HT maior FK, et HD minor OY. Erit igitur medietas HT ad HD maior quam FK ad OY, quare erit QH ad AH maior quam FK ad OY. 
◉ Moreover, AH:HD = HD:HT [by construction], so AH:HD = onehalf HD:onehalf HT. Hence, AH:HD = QH:onehalf HT, since QH = onehalf HD [by construction], so AH:QH = HD:onehalf HT, and so QH:AH = onehalf HT:HD. But onehalf HT > FK [since HT > 2FK by previous conclusions], and HD < OY. Accordingly, onehalf HT:HD > FK:OY, so QH:AH > FK:OY. 
◉ Amplius, linea AQ secat circulum EBD. Sit punctus sectionis Q, et ducatur linea DQ, que erit equidistans QH. Eritque proportio QH ad HA sicut QD ad DA, et ita QD ad DA maior quam FK ad OY. Sed FK ad OY sicut GB ad BA. Erit igitur maior QD ad DA quam BG ad BA, et ita QD maior BG, et arcus QD maior arcu GB. 
◉ Furthermore, line AQ cuts circle EBD [represented in figure 6.4.3, p. 99]. Let Q’ [in figure 6.4.3a, p. 100] be the point of intersection, and draw line DQ’, which will be parallel to [line] QH. So QH:HA = Q’D:DA [in similar triangles AQH and AQ’D], and so Q’D:DA > FK:OY. But FK:OY = GB:BA [by previous conclusions]. Therefore, Q’D:DA > BG:BA, so Q’D > BG [since DA = BA], and arc Q’D > arc GB.⁑ 
◉ Amplius, producatur AQ usque ad punctum S ut sit AS equalis AI, et ducatur linea SI, que erit equidistans QH, et erit proportio SI ad QH sicut IA ad AH. Sed supra positum est quod IA ad HA sicut TQ ad QH; erit igitur SI equalis TQ. 
◉ Extend AQ to point S so that AS = AI, and draw line SI, which will be parallel to [line] QH, and [so] SI:QH = IA:AH. But it was posited earlier that IA:AH = TQ:QH, so SI = TQ. 
◉ Amplius, mutetur figura ad evitandam linearum intricationem multiplicem et propter defectum litterarum ad distinctionem nominum. Cum ergo IA sit equalis linee quam diximus AS, fiat circulus secundum quantitatem ipsarum [FIGURE 6.4.3b, p. 306]. Et loco S ponamus nomen N, et producantur AG, AB usque ad hunc circulum, et sint ABC, AGR; loco littere Q ponamus F. Dictum est quoniam arcus DF maior est arcu BG. Sit arcus BM equalis arcui DF, et ducatur linea AMU et linee IM, NM, et linea QM, que producatur usque ad exteriorem circulum. Et cadat in punctum Z, et ducantur linee ZA, ZG. 
◉ Now because of the lack of letters for designating the key points, let us revise the diagram to avoid the excessive tangle of lines. Accordingly, since IA = the line we have designated as AS, construct circle [NRZ in figure 6.4.3b, p.101] according to their length [as radius]. Let us replace S with the letter N, let AG and AB be extended to [points R and C on the circumference of] this circle, let [the resulting lines] be ABC and AGR, and let us replace the letter Q’ with F. It has been claimed [earlier] that [arc] DF [formerly DQ’] > arc BG. Let arc BM = arc DF, and draw line AMU, as well as lines IM, NM, and [draw] line QM, and extend it to the outer circle. Let it fall to point Z, and draw lines ZA and ZG. 
◉ Cum arcus BM sit equalis arcui DF, addito communi, erit arcus MF equalis arcui DB. Erit angulus NAM equalis angulo IAB, et latera lateribus equalia; erit MN equalis IB. Et cum positum sit supra, AQ equale AH, erunt AQ, AM equalia HA, AB, et angulus angulo. Erit QM equalis HB, et erit angulus QMN equalis angulo HBI, quoniam duo eius latera duobus illius equalia. Et basis que est IH equalis basi NQ, et angulus NMU equalis angulo IBC. 
◉ Since arc BM = arc DF, then if common arc [DM] is added [to both], arc MF = arc DB. [Accordingly] angle NAM = angle IAB, the [corresponding] sides [NA and IA, and MA and BA, of triangles NAM and IAB] will be equal [and will contain equal angles, so the triangles will be equal, by Euclid, I.4], and [so] MN = IB. And since AQ was assumed earlier to be equal to AH [because they are radii of the same circle passing through Q and H, sides] AQ and AM [of triangle AQM] = [corresponding sides] HA and AB [of triangle HAB], and angle [QAM contained by sides AQ and AM in triangle AQM is equal] to angle [HAB contained by equal corresponding sides HA and AB in triangle HAB, so the two triangles are equal, by Euclid, I.4]. [Hence] QM = HB, and angle QMN = angle HBI, since both the sides containing them [i.e., QM and NM, and HB and IB, respectively] are equal [as concluded earlier]. [Accordingly] base IH = base NQ, and angle NMU = angle IBC. 
◉ Sed angulus IBC equalis angulo HBA, et angulus HBA equalis angulo QMA; erit angulus NMU equalis angulo QMA. Et quoniam QMZ est linea recta, ut posuimus, erit angulus QMA equalis angulo UMZ, quare punctus N refertur ad Z a puncto M, et locus ymaginis ipsius Q. Hoc tamen deest probationi ut pateat MZ totam esse extra circulum, quod sic patebit. 
◉ But angle IBC [which is IBZ2 in figure 6.4.3a] = angle HBA [by previous conclusions], and angle HBA = angle QMA, [so] angle NMU = angle QMA. And since QMZ is a straight line, as we stipulated, angle QMA = [vertical] angle UMZ, so [angle of incidence NMU = angle of reflection UMZ, and so the form of] point N is reflected to [point] Z from point M, and its imagelocation is [point] Q. This still falls short of a proof to show that all of [line of reflection] MZ lies outside the circle, which will be demonstrated as follows. 
◉ Palam quoniam contingens ducta a puncto B cadet inter I et H, et remotio puncti B a puncto H quanta est puncti M a puncto Q, et IH equalis NQ. Igitur contingens ducta a puncto M cadet inter N et Q. Igitur QM secat circulum, quare tota MZ extra circulum, et ita propositum. 
◉ It is clear that the tangent drawn from point B will fall between I and H [since HBA and IBC are acute, by previous conclusions], and [it is clear] that point B lies as far from point H as point M lies from point Q, and IH = NQ [from previous conclusions]. Therefore, the tangent drawn from point M will fall between N and Q. QM therefore intersects the circle, so all of MZ [lies] outside the circle, and so what we set out [to be proven is demonstrated].⁑ 
◉ Amplius, quoniam angulus NMU equalis angulo UMZ, erit arcus NU equalis arcui UZ. Erit angulus NAU equalis angulo UAZ. Sed iam patuit quod angulus NAU equalis est angulo IAC. Erit angulus IAC equalis angulo ZAU. 
◉ Furthermore, because angle NMU = angle UMZ, arc NU = arc UZ. [So] angle NAU = angle UAZ. But it has already been shown that angle NAU [which is the same as angle NAM] = angle IAC [which is the same as angle IAB]. [Hence] angle IAC = angle ZAU. 
◉ Angulus BAG aut erit equalis angulo GAM, aut minor, aut maior. Sit equalis. Si igitur ab angulo IAB subtrahatur angulus BAG, et ab angulo ZAU angulus MAG, remanebit angulus IAG equalis angulo ZAG. Erit IG equalis ZG, et triangulus triangulo, et erit angulus IGA equalis angulo ZGA. Restabit angulus IGR equalis angulo ZGR. Sed angulus IGR equalis angulo TGA. Erit angulus TGA equalis angulo ZGR. Si igitur TG producatur, veniet ad Z, quare TGZ linea recta. Igitur I a puncto G refertur ad Z, et locus ymaginis eius T. 
◉ Angle BAG will be equal to, less than, or greater than angle GAM.⁑ Let it be equal [as represented in figure 6.4.3c, p. 102]. Accordingly, if angle [C]BAG[R] is subtracted from angle IAB[C], and angle [U]MAG[R] from angle ZAU, angle IAG[R] will be left equal to angle ZAG[R]. [Thus], IG = ZG, triangle [IAG equals] triangle [ZAG], and angle IGA = angle ZGA. It will follow that angle IGR = angle ZGR. But angle IGR = angle TGA [from previous conclusions]. [So] angle TGA = angle ZGR. Hence, if TG is extended, it will reach [point] Z [on the circle passing through points I and N in figure 6.4.3c], so TGZ is a straight line. [The form of point] I is therefore reflected to [point] Z from point G, and [point] T is its imagelocation. 
◉ Sit ergo Z visus. Reflectentur ad ipsum duo puncta N, I a duobus punctis M, G, et loca ymaginum puncta T, Q. Igitur linea TQ erit ymago linee IN, et probatum est supra quoniam TQ equalis est IN, et ita potest accidere in hiis speculis ymaginem esse equalem rei vise. 
◉ So let Z be the center of sight. [The forms of] the two points N and I will be reflected to it from the two points M and G, and the [respective] imagelocations will be points T and Q. Thus, TQ will be the image of line IN, and it was proven earlier that TQ = IN [i.e., IS in figure 6.4.3a], and so it can happen in these kinds of mirrors that the image is the same size as the visible object.⁑ 
◉ Si vero angulus BAG fuerit maior angulo GAM, erit angulus ZAG maior angulo IAG. Sit angulus KAG equalis angulo IAG. Quoniam punctus K dimissior puncto Z, et punctus M dimissior G, linea KG secabit lineam ZM. Secet in puncto L, Igitur, existente visu in puncto L, refertur N ad ipsum a puncto M, et locus ymaginis Q; refertur ad ipsum I a puncto G, et locus ymaginis T, secundum priorem probationem. Et ita TQ ymago IN, quod est propositum. 
◉ If, however, angle BAG > angle GAM [in figure 6.4.3d, p. 103], then angle ZAG > angle IAG. Let angle KAG = angle IAG. Since point K is lower than [i.e., to the left of] point Z, and point M is lower than [i.e., to the left of point] G, line KG will intersect line ZM. Let it intersect at point L. Accordingly, if the center of sight lies at point L, [the form of point] N is reflected to it from point M, and [point] Q is its image location; [the form of point] I is reflected to it from point G, and [point] T is its imagelocation, according to the preceding proof. And so TQ is the image of IN, which is what was set out [to be proven].⁑ 
◉ Si vero angulus BAG fuerit minor angulo GAM, erit angulus ZAG minor angulo IAG. Sit angulus OAG equalis angulo IAG, et producatur linea OG. Palam quoniam I refertur ad O a puncto G. Linea OG aut secabit lineam ZMQ extra circulum speculi aut non. [4.57] Si secet extra, et in puncto sectionis fuerit visus, refertur ad ipsum duo puncta I, N, et loca ymaginum T, Q, et ita redit propositum. 
◉ On the other hand, if angle BAG < angle GAM [as in figure 6.4.3e, p. 104], angle ZAG < angle IAG. Let angle OAG = angle IAG, and extend line OG. It is clear that [the form of point] I is reflected to [point] O from point G. Line OG will either intersect line ZMQ outside the [great] circle [FDGB] of the mirror, or it will not.⁑ 
◉ Si forsan linea OG secabit lineam ZMQ intra circulum, nec poterit aptari predicta probatio. Sed dico quoniam extra hanc totalem superficiem erit invenire punctum ad quod refertur duo puncta I, N a duobus punctis speculi, et ymago TQ. 
◉ If line OG should happen to intersect line ZMQ inside the circle [as represented in figure 6.4.3e, p. 104, where X is the intersectionpoint], the foregoing proof cannot be applied. Instead, I say that a point can be found outside that entire surface to which [the forms of] the two points I and N are reflected from two points [on the mirror], and [that] TQ [forms] the image. 
◉ Verbi gratia, palam quoniam angulus NAZ duplus ad angulum IAB, et angulus IAO duplus ad angulum IAG, secundum predicta. Et angulus NAZ non excedit angulum IAO in angulo maiori angulo NAI. Et duo anguli OAI, ZAN maiores tertio, quod est IAN, et duo OAI, IAN maiores tertio NAZ, et duo ZAN, NAI maiores tertio IAO. Habemus ergo tres angulos, quorum quilibet duo maiores tertio. 
◉ For instance, from the foregoing it is evident that angle NAZ is twice angle IAB [since angle IAB = angle NAM = angle MAZ, by construction], and angle IAO is twice angle IAG [since angle IAG = angle OAG, by construction]. Furthermore, angle NAZ exceeds angle IAO by an amount no greater than angle NAI.⁑ In addition, the two angles OAI and ZAN [together] are greater than the third [angle], which is IAN, the two [angles] OAI and IAN are greater than the third [angle] NAZ, and the two [angles] ZAN and NAI are greater than the third [angle] IAO. We therefore have three angles [IAO, IAN, and NAZ], any two of which [together] are greater than the third. 
◉ Igitur ex illis est facere angulum corporalem. Fiat angulus ille super A, et sit linea SA erecta super A, et angulus IAS sit equalis angulo IAO, angulus NAS equalis angulo NAZ. Angulus NAI manebit immotus, et fiet linea AS equalis lineis AN, AI, que omnes sunt equales. 
◉ From these [angles], therefore, a solid angle can be formed [by Euclid, XI.23].⁑ Form that angle at A [in figure 6.4.3f, p. 105], let line SA be erected at A, let angle IAS = angle IAO, [and let] angle NAS = angle NAZ. Angle NAI will remain where it is, and line AS will be formed equal to lines AN and AI, which are both equal.⁑ 
◉ Et producantur linee TS, QS. Palam quoniam angulus TAS equalis angulo TAO, et duo latera duobus lateribus. Erit basis TS equalis basi TO, et triangulus triangulo, et ita angulus GTA equalis angulo STA. Similiter, angulus QAS equalis angulo QAZ, et latera lateribus. Erit triangulus equalis triangulo, et angulus MQA equalis angulo SQA. 
◉ Then draw lines TS and QS. It is evident that angle TAS = angle TAO [by construction], and the two [corresponding] sides [TA and OA are equal] to the two [corresponding] sides [TA and AS]. [Hence] base TS = base TO, and triangle [TAS] = triangle [TAO], and so angle [O]GTA = angle STA [since G lies on TO]. Likewise, angle QAS = angle QAZ, and the [corresponding] sides [AS and AQ are equal] to the [corresponding] sides [AZ and AQ]. [So] triangle [QAS] = triangle [QAZ], and angle [Z]MQA = angle SQA [since M lies on line QZ].⁑ 
◉ Dividatur angulus TAS per equalia per lineam AY; sit Y punctus in quo linea illa secabit lineam TS. Palam, cum angulus IAG sit medietas anguli IAO, erit angulus TAG equalis angulo TAY, et angulus GTA equalis angulo YTA, et unum latus commune, scilicet TA. Erit TG equalis TY, et triangulus triangulo, et erit AY equalis AG, et ita Y in superficie spere. Igitur angulus IAG equalis angulo IAY, et latera lateribus. Erit triangulus triangulo equalis, et erit AGI equalis angulo AYI. Linea IY producta erit equalis IG. 
◉ Bisect angle TAS with line AY, and let Y be the point at which that line will intersect line TS. Since angle IAG is onehalf angle IAO, it is evident that angle TAG = angle TAY, whereas angle GTA = angle YTA, and one side, i.e., TA, is common [to both triangles TAG and TAY. Accordingly] TG = TY, triangle [TAG] = triangle [TAY], AY = AG, and so Y will lie on the surface of the sphere [from which the mirror is formed]. Thus, angle IAG = angle IAY, and the [corresponding] sides [IA and AG] = the [corresponding] sides [IA and AY. So] triangle [IAG] = triangle [IAY], [angle] AGI = angle AYI [and] line IY in its full extent = [line] IG.⁑ 
◉ Et producatur AY extra speram usque ad punctum P. Restabit angulus IGR equalis angulo IYP. Verum, cum TS sit equalis TO, et TY equalis TG, restat GO equalis YS. Igitur AY, YS equalia AG, GO, et basis AS equalis basi AO. Erit triangulus equalis triangulo; erit angulus AYS equalis angulo AGO. Restat angulus SYP equalis angulo OGR. Igitur duo anguli IGR, OGR equales sunt duobus angulis IYP, SYP. 
◉ Extend AY beyond the sphere to point P [in figures 6.4.3f and 6.4.3h, pp. 105 and 107]. Angle IGR will be left equal to angle IYP. But, since TS = TO, and TY = TG, it follows that GO = YS. Therefore, AY and YS are equal to AG and GO [respectively], and base AS = base AO. [Hence] triangle [AYS] = triangle [AGO], and [so] angle AYS = angle AGO. It follows that angle SYP = angle OGR. Thus, the two angles IGR and OGR are equal to the two angles IYP and SYP [respectively]. 
◉ Verum linea AS secat speram. Sit punctus sectionis O. Igitur tria puncta O, Y, D sunt in superficie spere, quare linea OYD est pars circuli spere, et est linea communis superficiei spere et superficiei ITASP, quare punctus I refertur ad punctum S a puncto Y, et locus ymaginis T. 
◉ But line AS intersects the sphere [of the mirror]. Let O’ [in figure 6.4.3h] be the point of intersection. Accordingly, the three points O’, Y, and D lie on the surface of the sphere, so line O’YD is a segment of a [great] circle of the sphere, and it is the common section of the sphere’s surface and plane ITASP, so [the form of] point I is reflected to point S from point Y [within plane of reflection ITASP], and T is the imagelocation. 
◉ Similiter, diviso angulo NAS per equalia per AZZ, probabitur predicto modo quoniam QZ equalis QM, et AZ equalis AM, et ZS equalis MZ, et duo anguli NZZ et SZZ equales duobus angulis NMU, ZMU. Et ita N refertur ad S a puncto Z, et locus ymaginis Q, et ita TQ ymago IN, quod est propositum. 
◉ Likewise, if angle NAS [in figure 6.4.3f. p. 105] is bisected by [line] AZ’Z«, it will be proven in the preceding way that QZ’ = QM, AZ’ = AM, Z’S = MZ, and the two angles NZ’Z« and SZ’Z« are equal to the two angles NMU and ZMU. And so [the form of point] N is reflected to [point] S from point Z’, and Q is the imagelocation, so TQ is the image of IN, which is what was set out [to be proven]. 
◉ Amplius, si a puncto I ducatur perpendicularis super NA, cadat inter N et Q, non extra N, cum angulus INA acutus, quoniam equalis angulo NIA, et si caderet perpendicularis illa extra N, esset acutus maior recto. Faciet ergo perpendicularis illa angulum rectum super NQ, quem angulum respiciet linea IN, quare linea IN maior illa perpendiculari, quare perpendicularis illa minor TQ. 
◉ Now if a perpendicular is dropped from point I to NA, it should fall between N and Q, not beyond N because angle INA is acute, since it is equal to angle NIA, and if that perpendicular were to fall beyond N, an acute [angle] would be greater than a right angle.⁑ Therefore, that perpendicular will form a right angle on NQ, and that angle will be subtended by line IN, so line IN > that perpendicular, and so that perpendicular < TQ [since TQ is equal to IN by the initial construction]. 
◉ Punctus linee NQ in quem cadit perpendicularis reflectitur ad punctum S, et ymago eius cadet in linea NA supra punctum Q, quia quanto remotiora sunt puncta que reflectuntur tanto loca ymaginum magis accedunt ad centrum circuli, ex decima quinti huius. 
◉ The [form of the] point on line NQ where the perpendicular falls is reflected to point S, and its image will lie on line NA above point Q because the farther away [from the mirror’s surface] the points that are reflected lie, the more their imagelocations approach the center of the circle, according to the tenth [proposition] of the fifth book.⁑ 
◉ Et quecumque linea ducatur a puncto T ad aliquod punctum NQ supra Q erit maior TQ. Igitur ymago perpendicularis erit maior ipsa perpendiculari. Eodem modo, quecumque linea ducatur a puncto I ad NQ inter hanc perpendicularem et IN, erit ymago ipsius maior ipsa. 
◉ Moreover, any line drawn from point T to any point on NQ above Q will be longer than TQ. Therefore, the image of the perpendicular will be longer than the perpendicular itself [which is shorter than IN]. By the same token, no matter what line is drawn to NQ from point I between this perpendicular and IN, its image will be longer than it. 
◉ Verum determinentur hec certius. Punctus N refertur ad Z a puncto M, et locus ymaginis Q. Linea QM secat circulum in puncto quod sit E. Contingens ergo ducta a puncto Z ad circulum cadet super punctum aliquod arcus ME, et contingens illa cadet supra Q, quoniam punctus in quem cadet erit finis contingentie et finis ymaginum, et puncta sub puncto illo qui est finis contingentie non poterunt reflecti; superiora poterunt. 
◉ But these claims may be determined more definitively [as follows. The form of] point N [in figure 6.4.3m, p. 108] is reflected to [point] Z from point M, and Q is the imagelocation. Line QM cuts the circle at point E. Therefore, the tangent drawn from point Z to the circle will fall at some point on arc ME, and that tangent will fall above Q, since the point where it will fall [on NQ] will form the endpoint of tangency [X on cathetus NA] and [thus] the limit of images, and points below that endpoint of tangency cannot be reflected, [whereas points] above it can.⁑ 
◉ Igitur perpendicularis ducta a puncto I, si ceciderit supra punctum qui est finis contingentie, refertur punctus in quem cadit, et erit ymago perpendicularis maior perpendiculari. Si vero perpendicularis cadat in punctum contingentie aut infra, non refertur punctus in quem cadit, quare nulla erit ymago perpendicularis. Verumptamen, quoniam finis contingentie est infra N, erunt inter finem contingentie et N infinita puncta quorum quodlibet reflectetur, et ymago cuiuslibet supra NQ. Et cuiuslibet linee ducte a puncto I ad aliquod illorum punctorum erit ymago maior linea cuius fuerit ymago. 
◉ Therefore, if the perpendicular dropped from point I falls above the endpoint [X] of tangency, the point where it falls is reflected, and the image of the perpendicular will be longer than the perpendicular [itself]. But if the perpendicular should fall at or below the endpoint of tangency, the point where it falls is not reflected, so there will be no image of the perpendicular. However, since the endpoint of tangency lies below N, there will be an infinitude of points between the endpoint of tangency and N, and any of them will be reflected, and the image of any of them [will lie] on NQ. And the image of any line drawn from point I to any of those points will be longer than the line of which it is the image. 
◉ Igitur accidit in hiis speculis ymaginem aliquando equalem rei vise, aliquando maiorem, quod erat explanandum. Huius autem rei explanationem nec scriptam legimus nec aliquem qui dixisset aut intellexisset audivimus 
◉ In these [sorts of] mirrors, then, the image may sometimes be the same size as the visible object and sometimes larger, which is what was set out to be explained. Moreover, we have not read an explanation of this matter in any text, nor have we heard anyone who has discussed it or thought about it. 
◉ Amplius, in hiis speculis linee recte videntur curve, ut in pluribus curvitate quidem speculum non respiciente sed ei adversa. Similiter, curve apparebunt in hiis speculis curve, et si curvitas speculum respexerit, contrario situ apparebit, et hoc quidem intelligendum non in omnibus sed in pluribus, ad cuius rei explanationem necesse est quedam antecedentia premittere, unum quorum est: 
◉ Moreover, in these [sorts of] mirrors straight lines appear curved, such that, in many cases, the curvature [of their images] does not correspond to that of the mirror but is opposite. Likewise, curved [things] will appear curved in these [sorts of] mirrors, and if the curvature corresponds to that of the mirror, it will appear in an opposite orientation, but this must be understood not [to hold] in all cases but in several, [but] for the sake of explaining this, certain preliminary points must be set out, one of them being as follows. 
◉ [PROPOSITIO 4] Si fuerint duo puncta eiusdem longitudinis a centro speculi et inequalis longitudinis a centro visus, ymago puncti remotioris a centro visus erit remotior a centro spere quam propinquioris, et finis contingentie remotioris remotior a centro fine contingentie propinquioris, sive puncta illa sint in eadem superficie cum centro visus, sive in diversis. 
◉ [PROPOSITION 4, LEMMA 1] If two points lie the same distance from the center of the mirror and different distances from the center of sight, the image of the point lying farther from the center of sight will lie farther from the center of the sphere [forming the mirror] than the [image of the point] lying nearer [the center of sight], and the endpoint of tangency for the farther [point will lie] farther from the center [of the circle] than the endpoint of tangency for the nearer [point, and this will be the case] whether those points lie in the same plane as the center of sight or in different planes. 
◉ Probatio: Sint T, D [FIGURE 6.4.4, p. 307] duo puncta equaliter a G centro speculi remota, E centrum visus. Superficies DGT secabit speculum super circulum qui sit AB. Et sit angulus EGD equalis angulo TGZ, angulus EGT equalis angulo TGH, et sumatur in circulo punctum a quo T refertur ad Z, quod sit Q. Dico quoniam T non refertur ad H ab aliquo puncto BQ. 
◉ The proof [is as follows]. Let T and D [in figure 6.4.4, p. 109] be two points equidistant from G, the center of the mirror, [and let] E be the center of sight. Plane DGT will cut the mirror along [great] circle AB. Let angle EGD = angle TGZ, [let] angle EGT = angle TGH, and find point Q on the circle from which [the form of point] T is reflected to [point] Z [by book 5, proposition 25, in Smith, Alhacen on the Principles, 427432].⁑ I say that [the form of point] T is not reflected to [point] H from any point on BQ. 
◉ Palam enim quoniam non a puncto Q. Si autem sumatur punctum quodcumque in BQ, linea ducta a puncto H ad illud punctum secabit lineam QZ. Igitur ad illud punctum sectionis refertur T a puncto sumpto in BQ, et ad idem punctum sectionis refertur a puncto Q. Igitur T refertur ad idem punctum a duobus punctis illius circuli, quod est impossibile in hiis speculis, ut in libro quinto patuit. 
◉ It is obvious that [it does] not [do so] from point Q [itself]. Moreover, if some point is taken on BQ, the line [of reflection] drawn to that point from point H will intersect line QZ. [The form of point] T is therefore reflected to that point of intersection from the point selected on BQ, and it is [also] reflected to that same point of intersection from point Q. So [the form of] point T is reflected to the same point from two points on that circle, which is impossible in these [sorts of] mirrors, as was shown in [proposition 16 of] the fifth book [in Smith, Alhacen on the Principles, 412414]. 
◉ Restat ut T reflectatur ad H ab aliquo puncto QA. Sit illud M, et a puncto M ducatur contingens circulum usque ad lineam GT, que sit MN. Erit N finis contingentie T respectu H. 
◉ It follows that [the form of point] T may be reflected to H from some point on QA. Let that [point] be M [as found by book 5, prop. 25], and from point M draw MN to line GT tangent to that circle. N will be the endpoint of tangency for T with respect to H [as center of sight].⁑ 
◉ Et a puncto Q ducatur contingens que sit QO, que quidem necessario cadet sub MN. Producatur ZQ usque dum cadat super GT in puncto C. Erit C locus ymaginis Z. Erit igitur proportio GT ad TO sicut GC ad CO. Igitur maior erit proportio GT ad TN quam GT ad TO. Ergo multo maior GT ad TN quam GC ad CN. Sit ergo GT ad TN sicut GL ad LN. Erit GL maior GC, et L locus ymaginis H. 
◉ Then from point Q draw tangent QO, which will necessarily lie below MN. Extend ZQ until it falls on GT at point C. C will be the imagelocation [of T] for Z [as center of sight]. Thus, GT:TO = GC:CO [by book 5, proposition 7, in Smith, Alhacen on the Principles, 404]. So GT:TN > GT:TO. A fortiori, then, GT:TN > GC:CN. Accordingly, let GT:TN = GL:LN. GL > GC, and L will be the imagelocation [of T] for [center of sight] H [according to book 5, prop. 7]. 
◉ Sint ergo linee HG, EG, ZG equales, GF equalis GC, GS equalis GO. Cum igitur angulus EGD sit equalis angulo TGZ, et remotio D a puncto E sicut Z a puncto T, erit ymago D respectu G tantum elevata in linea GD quantum ymago T in linea GT. Igitur ymago D in puncto F. Et similiter, finis contingentie D respectu E erit eiusdem altitudinis cuius est finis contingentie Z, quare finis contingentie D in puncto S. 
◉ So let lines HG, EG, and ZG be equal, [let] GF = GC, [and let] GS = GO. Therefore, since angle EGD = angle TGZ [by construction], and since D lies as far from point E as Z does from point T [given that DG = TG, and EG = ZG, by construction], the image of D with respect to G will lie as high on line GD as the image of T on line GT [with respect to G, by book 5, proposition 17, in Smith, Alhacen on the Principles, 414415]. Thus, the image of [point] D [with respect to G lies] at point F [since GF = GC, by construction]. Likewise, the endpoint of tangency for D with respect to E will lie at the same height as the endpoint of tangency [at point O] for Z, so the endpoint of tangency for D [lies] at point S [since GS = GO, by construction]. 
◉ Verum, quoniam angulus EGT equalis angulo TGH, et HG equalis EG, erit L ymago T respectu E, sicut est respectu H. Et N finis contingentie respectu E, quare ymago puncti remotioris ab E remotior a centro ymagine propinquioris, et finis contingentie remotioris remotior a centro fine propinquioris, quod erat propositum. 
◉ But since angle EGT = angle TGH [by construction], and since HG = EG [also by construction], L will be the image of T with respect to E, just as it is with respect to H. And N is the endpoint of tangency with respect to E, so the image [L] of the point farther from E [i.e., T] lies farther from the center than the image [F] of the nearer [point D], and the endpoint of tangency [N] for the farther [point T lies] farther from the center than the endpoint of tangency [S] for the nearer [point D], which was what was set out [to be proven]. 
◉ [PROPOSITIO 5] Amplius, proposita linea AB [FIGURE 6.4.5, p. 307], et divisa in punctis G, D ut sit proportio AB ad BD sicut AG ad GD, si a punctis sectionis ducantur tres linee concurrentes in punctum unum, scilicet GE, DE, BE, et a puncto A ducatur linea secans illas tres lineas, dico quoniam linea illa divisa erit secundum predictam proportionem. 
◉ [PROPOSITION 5, LEMMA 2] Furthermore, given line AB [in figure 6.4.5, p. 109] divided at points G and D such that AB:BD = AG:GD, I say that, if three lines, i.e., GE, DE, and BE, are drawn from the points of division to intersect at one point, and if a line is drawn from point A to intersect those three lines, that line will be cut according to the aforesaid proportion. 
◉ Probatio: Ducatur linea AT secans tria latera GE, DE, BE in tribus punctis Z, H, T. Dico quoniam proportio AT ad TH sicut AZ ad ZH. 
◉ The proof [is as follows]. Draw line AT to cut the three sides GE, DE, and BE at the three points Z, H, and T. I say that AT:TH = AZ:ZH. 
◉ Ducatur a puncto H equidistans AB, que sit HQ. Palam quoniam proportio AB ad BD constat ex proportionibus AB ad HQ et HQ ad BD. Sed quoniam QH equidistans AB, erit triangulus TQH similis triangulo BTA, et erit proportio AB ad QH sicut AT ad TH. Similiter, triangulus QEH similis triangulo BED. Igitur erit proportio QH ad BD sicut HE ad ED. Ergo proportio AB ad BD constat ex proportionibus AT ad TH et HE ad ED. 
◉ From point H draw HQ parallel to AB. It is clear that AB:BD is compounded from AB:HQ and HQ:BD [i.e., AB:BD = (AB:HQ)(HQ:BD)]. But since QH is parallel to AB, triangle TQH will be similar to triangle BTA, and [so] AB:QH = AT:TH. Likewise, triangle QEH is similar to triangle BED. Therefore, QH:BD = HE:ED. Hence, AB:BD is compounded of AT:TH [which = AB:QH] and HE:ED [which = QH:BD]. 
◉ Producatur QH usque cadat super EG in puncto M. Proportio AG ad GD constat ex proportionibus AG ad HM et HM ad GD. Sed cum angulus EMH sit equalis angulo ZGD, erit angulus HMZ equalis angulo ZGA, et erit triangulus AZG similis triangulo HZM, et erit proportio AZ ad ZH sicut AG ad HM. 
◉ Extend QH until it falls on EG at point M. AG:GD is compounded from AG:HM and HM:GD. But since angle EMH = [alternate] angle ZGD, then angle HMZ [adjacent to EMH] = angle ZGA [adjacent to ZGD], and [so] triangle AZG will be similar to triangle HZM, and AZ:ZH = AG:HM. 
◉ Sed triangulus HEM similis triangulo GED. Erit proportio HM ad DG sicut HE ad ED. Igitur proportio AG ad GD constat ex proportionibus AZ ad ZH et HE ad ED, et eadem est AG ad GD que est AB ad BD. Igitur illa eadem constat ex proportionibus AT ad TH et HE ad ED, et similiter constat ex proportionibus AZ ad ZH et HE ad ED. Igitur eadem est proportio AT ad TH que est AZ ad ZH, et ita propositum. 
◉ But triangle HEM is similar to triangle GED. [Hence] HM:DG = HE:ED. Accordingly, AG:GD is compounded from AZ:ZH and HE:ED, and AG:GD = AB:BD [by construction].⁑ Thus, that same [proportion AG:GD] is compounded from AT:TH and HE:ED, and it is likewise compounded from AZ:ZH and HE:ED. Hence [if we drop the common term HE:ED], AT:TH = AZ:ZH, and so what was set out [has been proven]. 
◉ Eadem erit probatio quecumque linea ducatur a puncto A secans illas lineas tres concurrentes. Et si ducantur alie tres linee a tribus punctis G, D, B ad aliud punctum quam E concurrentes, et a puncto A ducatur linea quecumque secans eas, dividetur secundum predictam proportionem. Et ita quocumque modo concurrant tres linee, et si tres linee EG, ED, EB producantur ultra tria puncta B, D, G ex alia parte, et a puncto A ducantur linee secantes eas ex illa alia parte, numquam ille linee dividantur secundum predictam proportionem. 
◉ The same proof will hold no matter what line is drawn from point A to intersect those three intersecting lines. And if three other lines are drawn from the three point G, D, and B to intersect at some point other than E, and if any line is drawn from A to intersect those [three lines], it will be cut according to the aforesaid ratio. And so, however the three lines may intersect, if the [resulting] three lines [represented by] EG, ED, and EB are extended beyond the three points B, D, and G, on the other side [away from the point of intersection], and if lines are drawn from point A to intersect them on that other side, those lines may never be cut according to the aforesaid ratio.⁑ 
◉ [PROPOSITIO 6] Amplius, data linea AB predicto modo divisa, si a puncto A ducatur alia linea, velut AT, que dividatur iuxta eandem proportionem, et a punctis divisionum AB ducantur linee ad puncta divisionum AT, que quidem non sit equidistans, dico quoniam ille tres concurrent in uno eodem puncto. 
◉ [PROPOSITION 6, LEMMA 3] Moreover, given line AB [in figure 6.4.6, p. 110] cut in the preceding way, if another line, such as AT, is drawn from point A so as to be cut according to the same ratio, and if lines are drawn from the points of division on AB to the points of division on AT, those lines not being parallel, I say that those three [lines] will intersect at the same point. 
◉ Probatio: Sit proportio AT ad TH sicut AZ ad ZH. BT, DH non sunt equidistantes; igitur concurrent in puncto quod sit E. Linea GZ aut concurret ad idem punctum, aut non. Si ad illud, habemus propositum. Si non, ducatur linea EG. Secabit quidem lineam AT in alio puncto quam Z. Sit illud punctum L. Erit igitur proportio AT ad TH sicut AL ad LH, iuxta priorem probationem. Sed positum est AT ad TH sicut AZ ad ZH, et ita impossibile. 
◉ The proof [is as follows]. Let AT:TH = AZ:ZH. BT and DH are not parallel, so they will intersect at point E. Line GZ will either intersect [them] at the same point, or it will not. If [it does intersect] at that point, we have what was set out [to be proven]. If not, then draw line EG. It will intersect line AT in some point other than Z. Let that point be L. Thus, AT:TH = AL:LH, according to the previous proof. But it has been supposed that AT:TH = AZ:ZH, and so it is impossible [for EG to intersect AT at some point other than Z]. 
◉ Similiter, si ponatur quod linea GZ concurrat cum DH ad punctum E, probabitur hoc modo quod linea BT concurret ad idem. Similiter, si ponatur quod GZ, BT concurrant ad punctum E, convincetur quod DH concurret ad idem. 
◉ Likewise, if it is supposed that line GZ intersects DH at point E, it will be proven in this way that line BT will intersect [it] at the same [point]. So too, if it is supposed that GZ and BT intersect at point E, it will be indisputable that DH will intersect at the same [point]. 
◉ [PROPOSITIO 7] Amplius, divisa AB secundum hanc proportionem, si fuerint linee GZ, DH, BT equidistantes, et ducatur AT dividens illas, erit AT divisa secundum hanc proportionem. 
◉ [PROPOSITION 7, LEMMA 4] Furthermore, given that AB [in figure 6.4.7,p. 110] is divided according to this ratio [AB:BD = AG:GD], if lines GZ, DH, and BT are parallel, and if AT is drawn to cut them, AT will be divided according to this ratio. 
◉ Probatio: Cum DH sit equidistans GZ, erit proportio AZ ad ZH sicut AG ad GD, et cum BT sit equidistans DH, erit AB ad BD sicut AT ad TH. Sed AB ad BD sicut AG ad GD; erit AT ad TH sicut AZ ad ZH, et ita propositum. Hiis premissis, accedamus ad propositum. 
◉ The proof [is as follows]. Since DH is parallel to GZ, AZ:ZH = AG:GD, and since BT is parallel to DH, AB:BD = AT:TH. But AB:BD = AG:GD, [so] AT:TH = AZ:ZH, and so [we have demonstrated] what was set out [to be proven]. With these points established, let us proceed to what was proposed [in paragraph 4.72, p. 175176 above]. 
◉ [PROPOSITIO 8] Et primum, de arcu declaretur quomodo in hiis speculis ymago eius sit curva curvitate quidem speculum non respiciente, sed centrum. 
◉ [PROPOSITION 8] First of all, it must be shown how in these [sorts of] mirrors the image of an arc is curved with a curvature that accords not with the [surface of the] mirror but with its center. 
◉ Verbi gratia, sit AB [FIGURE 6.4.8, p. 308] arcus oppositus speculo, et sit G centrum illius arcus et similiter centrum speculi, D centrum visus. Et ducantur linee DG, AG, BG. Et sumatur E in arcu AB quocumque modo, et ducatur linea EG. Linea DG non sit in superficie ABG. Linea DG aut erit ortogonalis super superficiem ABG, aut declinata. 
◉ For instance, let AB [in figure 6.4.8, p. 110] be an arc facing the mirror [composed from sphere YZ on whose surface Y’Z’ is a great circle within the plane of arc AB], let G be the center of that arc as well as of the mirror, [and let] D [be] the center of sight. Draw lines DG, AG, and BG. Take E at random on arc AB, and draw line EG. Let line DG not lie in plane ABG. Line DG will either be perpendicular to plane ABG, or [it will be] inclined [to it]. 
◉ Sit ortogonalis. Erunt anguli DGA, DGE, DGB equales, et latera lateribus, quare bases equales. Igitur omnia puncta arcus AB eiusdem longitudinis erunt a centro visus, quare ymagines omnium eiusdem longitudinis a centro. 
◉ Let it be perpendicular. Angles DGA, DGE, and DGB will be equal, and the [corresponding] sides [of triangles DGA, DGE, and DGB will be equal] to the [corresponding] sides, so the bases [DA, DE, and DB will be] equal. Hence, all the points on arc AB will lie the same distance from the center of sight, so the images of all of them [will lie] the same distance from the center. 
◉ Sint Q, M, L ymagines A, E, B. Erit igitur GQ equalis GM et GL, quare QML erit arcus, et convexitas ipsius respectu centri non respectu speculi, sive reflexionis loci, quod est propositum. 
◉ Let Q, M, and L be the images of A, E, and B. Accordingly, GQ will be equal to GM and GL, so QML will be an arc, and its convex curvature accords with the center [of curvature], not with the [surface of the] mirror, or with the points of reflection, which is what was set out [to be proven].⁑ 
◉ Si vero linea DG non fuerit perpendicularis super superficiem AGB, ducta perpendiculari a puncto D super hanc superficiem, cum illa perpendicularis sit minor omnibus lineis ductis a puncto D ad hanc superficiem, erit angulus quem respicit hec perpendicularis supra G minor quolibet angulo supra punctum G intellecto quem respiciat alia linea a puncto D ad superficiem ducta, et linea ducta a puncto D ad superficiem quanto remotior erit a perpendiculari, tanto maior erit, et respiciet maiorem angulum. Si igitur hec perpendicularis non cadat in arcu AEB, sed ex parte una, erunt omnes linee ducte a puncto D ad hunc arcum declinate in partem unam, et remotiores maiores et maiorem angulum respicientes. 
◉ If, however, line DG is not perpendicular to plane AGB [as in figure 6.4.8a, p. 111], and if a perpendicular [DX] is dropped from point D to that plane, then, since that perpendicular is the shortest of all [possible] lines extending from point D to this plane, the angle this perpendicular forms with respect to G will be smaller than any angle imagined at point G formed by any other line drawn from point D to the plane, and the farther the line drawn from point D to the plane will lie from the perpendicular, the longer it will be and the greater the angle it will form. Accordingly, if this perpendicular does not fall on arc AEB but on one side of it, all the lines drawn from point D to this arc will be slanted to one side, and the ones that lie farther away will be longer and will form a larger angle.⁑ 
◉ Sit ergo, et sumantur tria puncta in arcu, scilicet E, C, B. Finis contingentie puncti B sit L; finis contingentie puncti C sit M, quoniam, cum C propinquius D quam B, erit M propinquior G quam L, et ita CM maior BL. 
◉ Let [this] be [the case], then, and take three points, i.e., E, C, and B, on the arc [in figure 6.4.8b, p. 111].⁑ Let L be the endpoint of tangency for point B, and let M be the endpoint of tangency for point C, for, since C is nearer than B to D, M will be nearer than L to G, and so CM > BL.⁑ 
◉ Q sit ymago C, T ymago B, et ducatur TQ. Et ducantur linee CB, ML, que quidem producte concurrent, si enim a puncto M duceretur equidistans CB, secaret ex GB lineam equalem CM. Concurrant in puncto O. 
◉ Let Q be C’s image, let T be B’s image, and draw TQ. Then draw lines CB and ML, which will intersect when extended, for if a line were drawn from M parallel to CB, it would cut a line from GB equal to CM [and thus be shorter than GL, which means that angle MLG < angle CBG]. Let them intersect at point O. 
◉ Et quoniam proportio GC ad CM sicut GQ ad QM, similiter, BG ad BL sicut GT ad TL, linea QT concurret cum lineis CB, ML. Et sit concursus in puncto O. 
◉ Since GC:CM = GQ:QM [by book 5, prop. 7], and likewise, since BG:BL = GT:TL [by the same proposition], line QT will intersect lines CB and ML [all at the same point]. Let the intersection be at point O.⁑ 
◉ Finis contingentie puncti E sit N [FIGURE 6.4.8a, p. 308]. Quoniam punctus N dimissior est puncto M, erit EN maior CM. Ductis ergo lineis EC, NM, concurrent. Sit concursus in puncto P, et ducatur linea QP, et procedat donec cadat super EG in puncto F. Et ducatur linea TQ usque ad EG, et cadat in puncto K. 
◉ Let N [in figure 6.4.8b, p. 111] be the endpoint of tangency for point E. Since point N is lower than point M [by proposition 4, lemma 1], EN > CM. Hence, if they are extended, lines EC and NM will intersect. Let the intersection be at point P, draw line QP, and extend it until it falls upon EG at point F. Then extend line [O]TQ to EG, and let it fall at point K. 
◉ Palam quoniam K erit supra F. Verum, cum proportio GC ad CM sicut GQ ad QM, et a punctis divisionum ducantur tres linee concurrentes in aliam partem producte, secabunt lineam EG secundum predictam proportionem, quare proportio GE ad EN sicut GF ad FN. Sed N finis contingentie, quare F locus ymaginis. Igitur linea FQT erit ymago arcus ECB, et erit linea curva non recta, quoniam TQK est recta, et curvitas linee non ex parte speculi. 
◉ It is evident that K will lie above F. However, since GC:CM = GQ:QM, and since three lines [EP, NP, and FP] are drawn through the points of division to meet [at point P] when extended to the other side, they will cut line EG according to the previous ratio [by proposition 6, lemma 3], so GE:EN = GF:FN. But N is the endpoint of tangency [for E, by construction], so F is the imagelocation [for E]. Hence, line FQT will be the image of arc ECB, and it will be a curved line rather than a straight one, for TQK is straight [by construction], and the curvature of the [resulting] line [TQF] is not oriented with [that of] the mirror. 
◉ Similiter, si perpendicularis a puncto D cadat ex alia parte arcus, similis erit probatio. Si vero cadat perpendicularis in medio arcus AB, linee a puncto D ex diversis partibus ad arcum ducte equaliter distantes a perpendiculari erunt equales, et equales angulos respicient supra G. Et ymagines earum equaliter a G distabunt, et finis contingentie similiter, et erit probare predicto modo de utraque parte arcus per se secundum quod dividitur a perpendiculari quod eius ymago sit linea curva modo predicto, quod est propositum. 
◉ Likewise, if the perpendicular dropped from point D falls on the other side of the arc [i.e., to the left of perpendicular D’G], the proof will be identical. But if the perpendicular falls on the midpoint of arc AB, the lines drawn to the arc from point D to opposite sides and equidistant from the perpendicular will be equal and will form equal angles at G. Moreover, their images will be equidistant from G, and so will the endpoint[s] of tangency,46 and it can be proven in the foregoing way for either side of the arc by itself, according to how it is cut by the perpendicular, that its image is a curved line in the way prescribed, which is what was set out [to be proven]. 
◉ [PROPOSITIO 9] Amplius, sumatur circulus cuius non sit centrum centrum speculi; verumptamen, sit in eadem superficie cum centro speculi. Dico quoniam, si in hoc circulo exteriori sumatur arcus ex parte centri speculi, id est propinquior ei, erit eius ymago curva. 
◉ [PROPOSITION 9] Now take a circle [containing arc ADB in figure 6.4.9, p. 112] whose center [F] is not the center [G] of the mirror [containing arc HK]; nonetheless, let it lie in the same plane as the mirror’s center. I say that, if an arc [ADB] is taken on the [larger] outer circle on the side of the mirror’s center, i.e., nearer that center [G and thus directly opposite it], its image will be curved. 
◉ Dato enim hoc arcu, ducatur linea a centro speculi ad centrum exterioris circuli, et producatur hec linea usque ad arcum datum. Linea ducta a centro speculi ad hunc arcum, que est pars dyametri maioris circuli, erit brevior omnibus lineis ductis ab eodem centro speculi ad illum arcum. Et a centro speculi possunt duci ad arcum datum due equales a diversis partibus huius brevis, que quidem maiores illa brevi. Et si secundum alteram illarum fiat circulus cuius centrum centrum speculi, transibit per capita harum duarum linearum arcus excedens arcum datum. 
◉ For, given this arc, draw a line [FG] from the center of the mirror to the center of the outer circle, and extend this line to the given arc [ADB]. The line drawn from the center of the mirror to this arc [i.e., GD], which is a segment of the diameter [FD] of the larger circle, will be shorter than all the [other] lines drawn from the same centerpoint of the mirror to that arc. Moreover, two equal lines [GA and GB] can be drawn to the given arc from the mirror’s centerpoint on opposite sides of this shortest line, and they will of course be longer [than it]. And if a circle is drawn according [to the length of] either of them from the center of the mirror, arc [AEB on it] will pass through the endpoints of these two lines, and it will be longer [and of a sharper curvature] than the given arc [ADB]. 
◉ Et palam quod ymago huius arcus excedentis erit linea curva, secundum predicta. Et ymagines punctorum huic arcui et arcui dato communium eedem, et medius punctus arcus excedentis remotior a centro puncto arcus dati quod ipsum respicit, quare eius ymago propinquior centro quam ymago puncti arcus dati illum respicientis. Et ita cuiuslibet puncti arcus exterioris ymago propinquior centro ymagine puncti arcus dati quod illud respicit. Quare ymago arcus dati curvior quam ymago arcus exterioris, quare ymago arcus dati curva est, quod est propositum. 
◉ It is clear that the image of this longer arc will be a curved line, according to previous conclusions [in proposition 8]. [It is] also [clear that] the images of the points [A and B] common to this arc [AEB] and the given arc [ADB will be] the same and [that] the midpoint [E] of the longer arc lies farther from centerpoint [G] than the [mid]point [D] on the given arc that corresponds to it, so its image [will lie] nearer to the centerpoint [G] than the image of point [D] on the given arc that corresponds to it [by book 5, prop. 17, in Smith, Alhacen on the Principles, 414415]. And so, the image of any point on the outer arc [will lie] nearer the centerpoint than the image of the point on the given arc that corresponds to it. Accordingly, the image of the given [less sharply curved] arc [ADB] is more sharply curved than the image of the [more sharply curved] outer arc [AEB], so the image of the given arc is curved, which is what was set out [to be proven]. 
◉ [PROPOSITIO 10] Amplius, quod linee recte ymago in hiis speculis sit curva sic probatur. 
◉ [PROPOSITION 10] In addition, it is proven as follows that the image of a straight line is curved in these [sorts of] mirrors. 
◉ Sit AB [FIGURE 6.4.10, p. 309] linea visa, G centrum speculi. Ducantur linee AG, BG. Aut sunt equales, aut non. Si equales, fiat circulus cuius G centrum ad quantitatem illarum, qui sit AEB. Cadet quidem linea AB intra circulum. Palam ex predictis quoniam ymago arcus AEB erit curva. Sit ergo ymago eius ZTH. Ymago A sit Z; ymago B sit H; ymago E sit T. 
◉ Let A[C]B [in figure 6.4.10, p. 112] be the [straight] line that is seen, [and let] G [be] the center of the mirror. Draw lines AG and BG. They are either equal or not. If [they are] equal, then construct [the] circle [containing arc] AEB on centerpoint G according to their length. Line A[C]B will obviously fall inside [this] circle. From the previous [proposition] it is clear that the image of arc AEB will be curved. Accordingly, let its image be ZTH. Let Z be the image of A, let H be the image of B, and let T be the image of E. 
◉ Et ducatur linea GE secans AB in puncto C. Palam quoniam E est in eadem linea cum C remotior a centro quam C. Erit eius ymago propinquior centro quam ymago C. Sit ergo M. Palam ergo quod linea ZMH est ymago linee AB, et est linea curva, quod est propositum. 
◉ Extend line GE to cut AB at point C. It is clear that E lies on the same line with C [and] farther from the centerpoint [G] than C. Its image will [therefore] lie nearer to centerpoint [G] than the image of C [by book 5, prop. 17]. So let it be M. It is clear that line ZMH is the image of line AB, and it is a curved line, which is what was set out [to be proven]. 
◉ [PROPOSITIO 11] Si vero linee AG, BG fuerint inequales, linea AB protracta aut secabit speculum, aut non. Sit quod non secet, et sit AG maior BG [FIGURE 6.4.11, p. 309], et fiat circulus supra G ad quantitatem AG, qui sit AEQ, et producatur AB usque cadat in circulum ex parte B. Cadat in puncto Q. 
◉ [PROPOSITION 11] On the other hand, if lines AG and BG are not equal, then, when it is extended, line AB will either intersect the mirror or not. Let it not intersect [as in figure 6.4.11, p. 113], let AG > BG, construct circle AEQ at [centerpoint] G according to the length of AG [as radius], and extend AB until it touches the circle on the side of B. Let it fall at point Q. 
◉ Patet ex superioribus quoniam ymago arcus AE est curva. Punctus ymaginis A sit Z; punctus ymaginis E sit M. Erit ZM ymago arcus AE, et quoniam ymago puncti B remotior a centro quam ymago puncti E, erit ymago linee AB curva, quod per puncta media arcus AE et linee AB poterit probari, quod est propositum. 
◉ It is clear from the foregoing [analyses in propositions 8 and 9] that the image of arc AE is curved. Let Z be the imagepoint for A, and let M be the imagepoint for E. ZM will [therefore] be the image of arc AE, and since the image of point B lies farther from centerpoint [G] than the image of point E [by book 5, prop. 17], the image [TNZ] of line AB will be curved, [and] this can be demonstrated according to the midpoints of arc AE and line AB, which is what was set out [to be proven]. 
◉ Nota quod in priori figura, si secetur a linea AB ex parte A pars quedam, et ex parte B secetur pars ei equalis, residuum linee habebit ymaginem curvam, et erit eadem probatio que est de linea AB. Et in hac figura secta alia parte linee AB ex parte B, de residuo erit eadem probatio que est de linea AB. 
◉ Note that in the preceding figure, if a segment is cut from line AB on the side of A, and a segment is cut on the side of B equal to it, the remaining portion of the line will have a curved image, and the proof [of this] will be the same as it is for [the whole of] line AB. Also, if in this figure another segment of line AB is cut on the side of B, the same proof will hold for the remainder [of the line] as holds for [the whole of] line AB. 
◉ [PROPOSITIO 12] Si vero linea AB tangit speculum, aut secabit, aut continget. Sit contingens [FIGURE 6.4.12, p. 309]. G sit centrum speculi, et ducantur linee AG, BG. Superficies ABG secat speculum supra circulum communem, qui sit SEZ. Palam quoniam linea AB continget speculum in hoc circulo. Contingat in puncto E. Protrahatur ergo AB usque ad E. D sit centrum visus. Superficies in qua sunt linee DG, AG secat speculum supra circulum communem superficiei et speculo. Et sit arcus illius circuli ZP. Similiter, linea communis superficiei in qua sunt DG, BG et circulus arcus illius circuli sit HP. 
◉ [PROPOSITION 12] But if line AB touches the mirror, it will either intersect it or be tangent to it. Let it be tangent [as in figure 6.4.12, p. 114]. Let G be the center of the mirror [containing arc PE in gray], and draw lines AG and BG. Plane ABG cuts the mirror along the circle SEZ [that forms their] common [section]. It is clear that line AB will be tangent to the mirror on this circle. Let it be tangent at point E. Accordingly, extend AB to E. Let D be the center of sight. The plane containing lines DG and AG cuts the mirror along a [great] circle [that forms the] common [section] of the plane and the mirror. Let ZP be an arc on that circle. Likewise, let HP be an arc on the [great] circle [that forms] a common section of the plane containing DG and BG [and the mirror].⁑ 
◉ Palam quoniam B refertur ad D ab aliquo puncto arcus HP. Si a puncto illo ducatur contingens, secabit lineam BG, et punctus sectionis erit finis contingentie. Sit punctus ille M. 
◉ It is clear [from proposition 4, lemma 1, paragraphs 4.7476] that [the form of point] B is reflected to [point] D from some point [F’] on arc HP. If a tangent is drawn from that point, it will intersect line BG, and the point of intersection will be the endpoint of tangency [for point B on cathetus BG]. Let M be that point [on the resulting tangent F’M]. 
◉ Palam etiam quod, si a puncto M ducatur contingens super circulum SEH, cadet contingens illa citra E, quoniam AB est contingens in puncto E, et punctus B est altior puncto M. Cadet ergo in puncto F, que contingens producta secabit lineam AE. Secet in puncto T. Ex alia parte secabit lineam AG. Secet in puncto C. 
◉ It is also clear that, if a tangent is drawn from point M to circle SEH, that tangent will fall in front of E because AB is tangent at point E, and point B lies above point M. It will therefore fall at point F, and, when it is extended, the tangent [MF] will intersect line AE. Let it intersect at point T. It will intersect line AG on the other side. Let it intersect at point C. 
◉ Fiat angulus BGS equalis angulo BGD, et producatur GS usque ad punctum L ad equalitatem linee DG. Erit ergo arcus HS equalis arcui HP, et sicut refertur B ad D a puncto arcus HP, refertur ad L a puncto arcus HS. Et erit reflexio a puncto F sicut in arcu HP est reflexio a puncto a quo ducatur contingens ad punctum M, et illa duo puncta a puncto M eiusdem longitudinis. Ducantur ergo linee BF, LF. 
◉ Form angle BGS = angle BGD, and extend GS to point L so that it is equal to line DG. Accordingly, arc HS = arc HP, and just as [the form of point] B is reflected to [point] D from a point on arc HP, it[s form] is reflected to L from some point on arc HS. Moreover, the reflection will occur from point F just as the reflection on arc HP occurs from the point [F’] from which the tangent [F’M’] is drawn to point M, and those two points lie the same distance from point M [which means that the reflections from B to D and from B to L are perfectly equivalent]. So draw lines BF and LF. 
◉ A refertur ad D ab aliquo puncto arcus ZP. Verum in triangulo HZP duo arcus HZ, HP sunt maiores tertio, scilicet ZP. Sed HP est equalis HS. Igitur ZP est minor ZS. Rescindatur ZS ad equalitatem in puncto Y, et ducatur linea GY, que producta ad equalitatem GD secabit necessario lineam FL. Secet in puncto X, et sit GXK equalis GD. 
◉ [The form of point] A is reflected to D from some point [R] on arc ZP. However, in triangle HZP the two arcs HZ and HP are longer than the third, i.e., ZP.⁑ But HP = HS. Therefore, ZP < ZS. Cut an equal segment from ZS at point Y [so that ZY = ZP], and draw line GY, which, when it is extended to the same length as GD, will necessarily intersect line FL. Let it intersect at point X, and let GXK = GD. 
◉ Palam quoniam sicut A refertur ad D ab aliquo puncto arcus ZP, similiter, refertur ad K ab aliquo puncto arcus ZY. Dico quoniam non refertur ad ipsum nisi a puncto quod est citra F ex parte Z. 
◉ It is clear that, just as [the form of point] A is reflected to D from some point [R] on arc ZP, it is likewise reflected to K from some point [R’] on arc ZY.⁑ I say that it is reflected to it [i.e., K] only from a point that is below F on the side of Z. 
◉ Si enim dicatur quod potest a puncto F vel aliquo puncto arcus FY, linea ducta a puncto A ad punctum reflexionis secabit lineam BF. Ad illud punctum sectionis reflectitur punctus K, et ad idem punctum refertur punctus L, et ita duo puncta in hiis speculis reflectuntur ad idem punctum ex eadem parte, quod est impossibile. Restat ut punctus A refertur ad K ab aliquo puncto arcus ZF. 
◉ For if it is claimed that it can [be reflected] from point F or from some point on arc FY, the line drawn from point A to the point of reflection will intersect line BF. [The form of] point K is reflected to that point of intersection, and [the form of] point L is reflected to the same point, and so two points are reflected in these [sorts of] mirrors to the same point on the same side, which is impossible [by book 5, prop. 16]. It follows that [the form of] point A is reflected to [point] K from some point [R’] on arc ZF. 
◉ Si ab illo puncto ducatur contingens, secabit lineam AZ, et cadet inter C et Z, quoniam punctus F dimissior quolibet puncto arcus ZF, et ita contingens a puncto F altior aliis a punctis arcus ZF ductis. Cadat ergo contingens illa in puncto N, et ducatur linea NM, que quidem linea, cum transeat per acumen trianguli BMT et producta dividat angulum, necessario secabit BT. Secet in puncto Q, et ducatur linea GQ. 
◉ If a tangent is drawn from that point, it will intersect line AZ, and it will fall between C and Z because point F is lower than any [other] point on arc ZF, so the tangent from point F is higher than the rest that are drawn from points on arc ZF. So let that tangent [R’N] fall at point N, and draw line NM, and since it passes through the vertex of triangle BMT and cuts the angle when extended, that line will necessarily intersect BT. Let it intersect at point Q, and draw line GQ. 
◉ Sit autem I ymago puncti A; O sit ymago puncti B; U sit ymago puncti Q. Palam, cum B sit propinquior puncto G quam A, erit O remotior a puncto G quam I. Ducatur ergo linea IO. Palam etiam quod proportio AG ad AN sicut GI ad IN, et proportio BG ad BM sicut GO ad OM. Cum ergo linee AG, BG dividantur secundum hanc proportionem utraque in tribus punctis, et a punctis divisionum ducantur linee quarum due, scilicet AB, MN, concurrant ad idem punctum, scilicet ad idem punctum Q, tertia necessario concurret ad idem illud punctum. 
◉ Now let I be the image of point A; let O be the image of point B; and let U be the image of point Q. Since B lies nearer than A to point G, O will lie farther than I from point G [by book 5, prop. 17]. So draw line IO. It is also clear [from book 5, prop. 7] that AG:AN = GI:IN, while BG:BM = GO:OM. Thus, since lines AG and BG are each cut in three points according to this ratio, and since two of the lines, i.e., AB and MN, extended from the points of division [A and N on base AG] intersect at the same point, i.e., at the same point Q, the third [line extending from point I on base AG] will necessarily intersect [these two] at that same point [by proposition 6, lemma 3]. 
◉ Igitur IO producta cadet supra Q, quare IOQ recta linea. Igitur IOU non erit recta. Sed IOU est ymago linee AQ, quare ymago linee AQ erit curva. Posito autem puncto B loco puncti Q, et aliquo puncto linee AB posito loco puncti B, erit eodem modo penitus probare quoniam ymago linee AB est curva, et hoc est propositum. 
◉ Therefore, when it is extended, IO will fall upon Q, so IOQ [forms] a straight line. Thus, IOU will not be a straight [line]. But IOU is the image of line AQ, so the image of line AQ will be curved. Furthermore, if point B replaces point Q, and if some point on line AB replaces point B, it will be demonstrable in exactly the same way that the image of line AB is curved, and this is what was set out [to be proven].⁑ 
◉ [PROPOSITIO 13] Si vero AB [FIGURE 6.4.13, p. 310] secat circulum, secet in puncto E, M finis contingentie linee BG. B refertur ad D ab aliquo puncto arcus HP. Arcus ab illo puncto reflexionis usque ad H aut est equalis arcui HE, aut maior, aut minor. 
◉ [PROPOSITION 13] If, however, AB intersects the circle [in figure 6.4.13, p. 116], let it intersect at point E, [and let] M [be] the endpoint of tangency on line BG. [The form of point] B is reflected to [point] D from some point [F’] on arc HP. The arc [extending] from that point of reflection to H [i.e., HF’] is either equal to, longer than, or shorter than arc HE. 
◉ Si equalis (sed palam quoniam arcus ille est equalis arcui HQ), sit Q punctus circuli in quem cadat contingens ducta a puncto M ex parte E. Igitur AE transit per punctum Q, et ita MQ secat AE per punctum E. 
◉ If it is equal (but it is evident that that arc is equal to arc HQ), let Q [in figure 6.4.13] be the point on the circle where the tangent drawn from point M falls on the side of E. Thus, AE passes through point Q, so MQ intersects AE through point E.⁑ 
◉ Si vero arcus ille minor est arcu HE, secabit MQ lineam AE ultra punctum Q, ut efficiatur triangulus EQT. 
◉ But if that arc [HF’] < arc HE [as in figure 6.4.13b, p. 118], MQ will intersect line AE beyond point Q [at point T] so as to form triangle EQT.⁑ 
◉ Si vero arcus ille fuerit maior arcu HE, secabit quidem linea MQ lineam AE citra punctum Q. 
◉ On the other hand, if that arc [HF’] > arc HE, line MQ will intersect line AE below point Q.⁑ 
◉ Sive hoc sive illud fuerit, iteretur predicta probatio, et eodem penitus modo probetur quoniam ymago linee AB est curva, quod est propositum. 
◉ Whether the latter or the former is the case, repeat the earlier proof, and it should be demonstrated in precisely the same way that the image of line AB is curved, which is what was set out [to be proven]. 
◉ [PROPOSITIO 14] Amplius, si in superficie in qua sunt linea visa et centrum spere fuerit centrum visus—superiora enim dicta sunt visu non existente in illa superficie—linea ergo visa recta aut concurret cum circulo communi illi superficiei et speculo, aut non concurret. 
◉ [PROPOSITION 14] Furthermore, if the center of sight lies in the plane containing the visible line and the center of the sphere—in the previous cases it was stipulated that the center of sight does not lie in that plane⁑—then the straight visible line will either intersect the [great] circle [forming] the common section of that plane and the mirror, or it will not intersect [it]. 
◉ Si concurret, aut erit perpendicularis super speculum aut declinata. Si perpendicularis, angulus illarum linearum cadet supra centrum speculi, que quidem linea videbitur recta, ymago enim cuiuslibet puncti illius linee apparebit in ipsa linea, et ita ymago illius linee recta. 
◉ If it will intersect, it will be either perpendicular to the mirror[‘s surface] or inclined to it. If [it is] perpendicular, the angle [formed by] those lines will fall on the center of the mirror, and [the image of] that line will appear straight, for the image of any point on that line will appear on that line, and so the image of that line [will be] straight. 
◉ Si vero linea proposita declinata fuerit, aut erit declinatio ex parte visus, aut ex alia parte. Si ex alia parte, sumatur punctus circuli a quo reflectatur aliquid ad visum, et sumatur linea reflexionis. Aliqua linearum declinatarum cadet forsitan super hanc lineam reflexionis, quod si fuerit, non videbitur quidem hec linea declinationis. 
◉ But if the given line [AB] is slanted, its slant will either be toward the center of sight or away from it. If it slants away from it [as in figure 6.4.14, p. 120],⁑ find the point [R] on the circle from which [the form of] some [point on it, such as B’] is reflected to the center of sight [D], and find the line of reflection [RD]. Any of the slanted lines may fall on this line of reflection, and if it does, then [an image of] this slanted line will not be seen. 
◉ Protracta a centro visus ad centrum speculi linea, si sumatur in arcu circuli citra hanc lineam punctus a quo refertur ad visum aliquis punctus linee declinationis. Sed ille punctus refertur a puncto prius assignato, qui est terminus linee reflexionis, cum linea declinationis sit supra lineam reflexionis, et ita ille punctus linee declinationis refertur ad visum a duobus punctis arcus, quod est impossibile. 
◉ Having extended a line [DG] from the center of sight to the center of the mirror, take a point [R’] on the arc of the circle in front of this line, such that [the form of] some point [B’] on the slanted line is reflected from it to the center of sight. But [the form of] that point is reflected from the previously designated point [R], which is the endpoint of the line of reflection, since the slanted line lies upon the line of reflection, and so [the form of] that point on the slanted line is reflected to the center of sight from two points on the arc, which is impossible [by book 5, prop. 16]. 
◉ Licet autem reflectatur punctus ille a puncto primum sumpto, non tamen videtur, cum sit in linea reflexionis, quoniam occultatur per precedentia puncta, et ita linea adiacens linee reflexionis non videtur. 
◉ Moreover, even though [the form of] that point may be reflected from the point that is initially selected, it[s image] is still not seen, since it lies on the line of reflection, so it[s image] is occluded by points in front of it [on the objectline], and so [the image of] a line lying upon the line of reflection is not seen.⁑ 
◉ Si vero sumatur linea declinationis cuius declinatio non ex parte visus iacens quidem sub linea reflexionis et secans ipsam in puncto circuli, dico quod nullus punctus illius linee videbitur. 
◉ But if a slanted line [AB in figure 6.4.14a, p. 120] is taken with its slant not toward the center of sight, and if it lies below the line of reflection [AB] and cuts it at a point [B] on the circle, I say that no [image of any] point on that line will be seen. 
◉ Sumpto enim puncto, si dicatur quod punctus ille potest reflecti ab aliquo puncto arcus interiacentis lineam reflexionis et lineam a centro visus ad centrum speculi ductam, et ducatur linea ab illo puncto ad punctum arcus sumptum, hec secabit lineam reflexionis, et punctus sectionis reflectitur ad visum a duobus punctis arcus, quod est impossibile. 
◉ For, given [such a] point [e.g., A], if it is claimed that [the form of] that point can be reflected from some point [R’] on the arc lying between the line of reflection [DB] and line [DG] extended from the center of sight to the center of the mirror, and if a line [of incidence AR’] is extended from that point to the point chosen on the arc, this line will intersect the line of reflection [BD], and the point of intersection [X] is reflected to the center of sight from two points on the arc, which is impossible.⁑ 
◉ Si vero dicatur quod punctus sumptus in linea refertur a puncto arcus circuli qui est sub ipsa linea, erit impossibile, quia ille totus arcus occultatur a linea. 
◉ On the other hand, if it is claimed that the [form of the] point [A] taken on the [slanted] line is reflected from a point on the arc of the circle below that line [i.e., to the right of B], it will be impossible [for it to be seen], since that whole arc is occluded by the line. 
◉ Si vero linea sumpta non attingit circulum, poterit quidem videri, sed modicum est. Si vero sumatur linea declinationis predicte inter lineam reflexionis et lineam per punctum reflexionis primo sumptum transeuntem ad centrum, poterit quidem videri hec linea, et minuetur curvitas ymaginis huius linee secundum quod magis accesserit ad lineam transeuntem ad centrum per punctum reflexionis. 
◉ If, however, the chosen line does not reach the circle, it can indeed be seen, but it is quite small. But if a line with the previous slant [i.e., away from the eye] is selected between the line of reflection and the line initially assumed to pass through the point of reflection to the center [of the circle], this line can in fact be seen, and the curvature of the image of this line will decrease as it approaches the line passing through the point of reflection to the center [of the circle].⁑ 
◉ Si vero sumantur linee inter lineam ad centrum transeuntem per punctum reflexionis, videbuntur quidem sive declinatio earum sit ex parte visus, sive non. Et modus visus earum simili modo visus linearum inter lineam reflexionis et lineam ad centrum transeuntem. Et hec quidem intelligenda sunt de lineis concurrentibus in arcu circuli qui apparet visui, id est in arcu qui interiacet duas contingentes ductas a centro visus ad circulum. 
◉ But if lines are chosen between the line passing through the point of reflection to the center [of the circle and the mirror], they will appear, whether their slant lies toward the center of sight or not. And the way they [i.e., lines slanting toward the eye] are seen is like the way lines [slanting away from the eye] between the line of reflection and the line passing to the center are seen. But these things must be understood for lines that meet the arc of the visible part of the circle, i.e., on the arc lying between the two [lines] drawn tangent to the circle from the center of sight. 
◉ Linearum autem concurrentium cum circulo in parte circuli occulta visui aliqua erit equidistans linee reflexionis. Illa quidem non videbitur. Similiter, conterminabilis equidistanti que est sub equidistanti occultabitur, sed conterminabilis equidistanti supra ipsam existens poterit videri. 
◉ On the other hand, among lines that meet the circle on the side of the circle that is invisible, one of them will be parallel to the line of reflection. That one will not be seen. Likewise, any one that borders on the parallel and lies below it will be invisible, whereas one that borders on the parallel [and lies] above it can be seen.⁑ 
◉ Si vero sumatur linea inter equidistantes non conterminabilis aliqui earum, si fuerit eius declinatio ex parte visus, videbitur. Si ex alia parte, aliquando videbitur, aliquando non, quoniam, si a termino eius producatur equidistans linee reflexionis, si fuerit linea illa sub equidistantem, non videbitur; si supra eam videri poterit. 
◉ If, however, a line is selected between the parallels but not bordering on any of them, and if it is slanted toward the center of sight, it will be seen. If it slants in the other direction, it will sometimes be seen, and sometimes not because, if a parallel to the line of reflection is extended from its endpoint, and if that line lies below the parallel, then it will not be seen, whereas [if it lies] above it can be seen.⁑ 
◉ Si vero linee non concurrant cum circulo, aut secabunt lineam ductam a centro visus ad centrum speculi, aut equidistabunt ei. Si secet aliquam earum, linea illa aut secabit eam ex parte visus, id est inter visum et speculum, aut ultra speculum. Si ultra, occultabitur linea illa, sed forsan apparebunt eius capita. Si vero secet lineam visualem ex parte visus, apparebit quidem similiter. Si fuerit equidistans linee visuali, poterit videri. Omnium autem harum linearum ymagines curve. 
◉ But if the lines do not meet the circle, they will either intersect the line drawn from the center of sight to the center of the mirror, or they will be parallel to it. If any of them intersects it, that line will either intersect it on the side of the center of sight, i.e., between the center of sight and the mirror, or [it will intersect it] beyond the mirror [and the center of sight]. If [it intersects] beyond [i.e., above the head], that line will not be visible, but its ends may appear. If it cuts the visual axis on the side of the center of sight, it will in fact appear the same [i.e., not visible in the mirror]. If it is parallel to the visual axis, it can be seen.⁑ Moreover, the images of all these lines are curved. 
◉ Visu autem existente in eadem superficie cum centro speculi et lineis visis, diminuta est apparentia, et que sic manifestius apparet est illa que declinata est maxima declinatione et illa visum respiciente. Pari modo, arcuum in hiis speculis apparentium et in eadem superficie cum centro speculi et visu existentium, ymagines quidem curve curvitate speculum respiciente. [4.140] Hec autem intelligenda sunt duplici visu existente in eadem superficie cum centro speculi et re visa. Si enim alter visus modicum declinetur quoad ipsum, alio modo res visa comprehendetur. Et visu existente extra superficiem rei vise et centri speculi, certior erit ipsius rei comprehensio quam existente in ea. 
◉ And if the center of sight lies in the same plane as the center of the mirror and the visible lines, they appear diminished, and the one that appears most clear in this case is the one that is most slanted and that corresponds to the center of sight. By the same token, the images of arcs that appear in these [sorts of] mirrors and that lie in the same plane as the center of the mirror and the center of sight appear curved according to the curvature of the mirror.⁑ 
◉ [PROPOSITIO 15] Quod autem ymago rei vise sit curva, visu existente in superficie centri speculi et rei vise, probabitur. 
◉ [PROPOSITION 15] That the image of a visible object is curved when the center of sight lies in the plane containing the mirror’s center and the visible object will be proven [as follows]. 
◉ Sit D [FIGURE 6.4.15, p. 311] centrum visus, G centrum speculi. HE sit linea visa, que quidem HE non concurrat cum circulo, sed sit equidistans linee DG, vel secet eam ex parte D. Sumatur superficies in qua sunt linea DG et linea HE; circulus communis huic superficiei et speculo sit AB. 
◉ Let D be the center of sight and G the center of the mirror. Let HE be the visible line. Let HE not intersect the circle but be parallel to line DG [as in figure 6.4.15, p. 122], or let it intersect it on the side of D [as in figure 6.4.15a, p. 122]. Take the plane containing line DG and line HE, and let circle AB be the common section of this plane and the mirror. 
◉ Producatur linea HG. Z sit ymago H, punctus circuli a quo refertur H ad D sit B, et a puncto B ducatur contingens, que secet lineam HG super punctum T. Erit T finis contingentie. 
◉ Draw line HG. Let Z be the image of H, let B be the point on the circle from which [the form of point] H is reflected to [point] D, let the tangent be drawn from point B, and let it intersect line HG at point T. T will be the endpoint of tangency [on cathetus HG]. 
◉ Ducatur linea GB, que producta necessario concurret cum HE, si enim HE fuerit equidistans DG, concurret quidem. Si vero DG concurrat cum HE, multo fortius GB concurret cum eadem. Concursus ille aut erit in linea HE, aut ultra hanc lineam. 
◉ Draw line GB, which will necessarily intersect HE when it is extended, for, if HE is parallel to DG, it will necessarily intersect. If, however, DG intersects HE, then a fortiori GB will intersect it. That intersection will lie either on line HE, or beyond that line. 
◉ Sit ultra. Concurrat in puncto M; ymago puncti M sit Q; finis contingentie sit S. Et ducatur linea ZQ, similiter linea TS, et producatur a puncto A contingens AU. Palam quoniam AB est minor quarta, quare D videat ex circulo minus medietate, quare angulus AGB est acutus, et angulus UAG est rectus. Igitur AU concurret cum GB. Concurrat in puncto U. Dico quoniam punctus U cadat supra punctum S. 
◉ Let it lie beyond. Let it intersect at point M, let Q be the image of point M, and let S be the endpoint of tangency [on cathetus MG]. Draw line ZQ, as well as line TS, and draw tangent AU from point A. It is clear that [arc] AB is less than onefourth [the circumference of the circle, since GD lies on a diameter of the circle and DB intersects the circle], so [the eye at point] D should see less than half the circle [when the corresponding arc below A is included], [and] so angle AGB is acute, while angle UAG is right. Hence, AU will intersect GB. Let it intersect at point U. I say that point U should fall above point S. 
◉ Cum enim punctus M reflectatur ab aliquo puncto arcus AB, et A sit dimissior illo puncto, erit finis contingentie A altior fine contingentie illius puncti. Et ita S dimissior puncto U. Procedat ergo TS donec concurrat cum linea AU, et sit concursus in puncto K. 
◉ For, since [the form of] point M is reflected from some point [X] on arc AB, and since A lies below that point, the endpoint of tangency for A [as a point of reflection for the form of any point on cathetus GM] will lie higher than the endpoint of tangency for that point [X as a point of reflection for any point on cathetus GM]. And so S [lies] below point U. Accordingly, extend TS until it intersects line AU, and let the intersection be at point K. 
◉ Et ducatur linea GK, que producta concurrat cum HM in puncto C. Punctus C refertur ad D ab aliquo puncto arcus AB. Sit ille punctus F, a quo ducatur linea contingens usque ad GC, que quidem dimissior linea AK, et erit punctus O dimissior puncto K. 
◉ Draw line GK, and let it intersect HM at point C when it is extended. [The form of] point C is reflected to [point] D from some point on arc AB. Let F be that point, and from it draw a tangent to GC, that tangent being lower than line AK, and [any] point [on it, such as] O will be lower than point K. 
◉ Sit O finis contingentie. Ducatur linea DF usque cadat super GC. Sit casus in puncto R. Et producatur ZQ usque ad lineam GC, et cadat in puncto L. Dico quoniam L est supra R. 
◉ Let O be the endpoint of tangency. Extend line DF until it falls on GC. Let it fall at point R. Extend ZQ to line GC, and let it fall at point L. I say that L lies above R. 
◉ Linee enim HC, TK, ZL aut sunt equidistantes, aut concurrent. Sint equidistantes. Cum ergo hee equidistantes, secent lineam CG super tria puncta C, K, L, et secent utramque linearum MG, HG. Et proportio HG ad HT sicut GZ ad ZT; similiter, MG ad MS sicut GQ ad QS. Erit proportio eadem GC ad CK sicut LG ad LK. 
◉ For either lines HC, TK, and ZL are parallel, or they will intersect. Let them be parallel [as in figure 6.4.15a]. Accordingly, since they are parallel, let them intersect line CG at the three points C, K, and L, and let them intersect both lines MG and HG. HG:HT = GZ:ZT [by book 5, prop. 7], and likewise MG:MS = GQ:QS [because HG and GM are cut equiproportionally by parallels HMC, TSK, and ZQL, and for that same reason] GC:CK = LG:LK [all according to proposition 7, lemma 4]. 
◉ Sed palam quoniam R est ymago C, linea enim DF linea reflexionis concurrens cum CG in puncto R, et O finis contingentie, quare proportio GC ad CO sicut GR ad RO. Sed maior GC ad CK quam GC ad CO, et ita maior GL ad LK quam GR ad RO. Ergo maior OR ad RG quam KL ad LG, et ita maior OG ad RG quam KG ad LG. Sed KG maior OG, quare LG maior RG. Igitur R dimissior puncto L. Sed ZQL est linea recta. Igitur ZQR est linea curva, et ita ymago linee HC est curva. Posito ergo aliquo puncto linee HC loco puncti M et puncto E loco puncti C, erit probare quod ymago HE est curva. 
◉ But it is clear that R is the image of C because line of reflection DF intersects CG at point R, and O is the endpoint of tangency, so GC:CO = GR:RO [by book 5, prop. 7]. However, GC:CK [which = GL:LK] > GC:CO [which = GR:RO], and so GL:LK > GR:RO. Accordingly, OR:RG > KL:LG, and so OG:RG > KG:LG [by Euclid, V.18]. But KG > OG, so LG > RG. Hence, R lies lower than point L. But ZQL is a straight line. Therefore, ZQR is a curved line, and so the image of line HC is curved. So if some point on line HC replaces point M, and if point E replaces point C, it will be demonstrable that the image of HE is curved. 
◉ Si vero linee HC, TS, ZQ concurrant, aut erit concursus ex parte D, aut ex parte HG. Sit ex parte D [FIGURE 6.4.15a, p. 311], et sit concursus in puncto C. Erit ZQC linea recta, quare ZQR erit curva, et ita ymago linee HE curva, quod est propositum. 
◉ But if lines HC, TS, and ZQ intersect, the intersection will either be on the side of D, or [it will be] on the side of HG. Let it be on the side of D [as represented in figure 6.4.15b, p. 123], and let the intersection be at point C. ZQC will be a straight line, so ZQR will be curved, and so the image of line HE [will be] curved, which is what was set out [to be proven].⁑ 
◉ Si vero proponatur arcus extra speculum, erit de eo probare quod ymago sit curva sicut probatum est visu non existente in eadem superficie cum arcu et centro speculi, et hoc est propositum. 
◉ If an arc is posed outside the mirror, however, it will be possible from this to prove that its image is curved just as it was proven [in proposition 11] when the center of sight did not lie in the same plane as the arc and the center of the mirror, and this is what was set out [to be proven]. 
◉ Igitur in hiis speculis linee recte apparent curve, et curve similiter apparent curve. Si autem proponatur visui in hiis speculis corpus curvum sed longum, modicum habens latitudinis, apparebit quidem illius corporis curvitas manifeste, cum ipsa discerni possit per ea que supra corpus aut intra. Non enim plane discernitur curvitas nisi magna, ubi occulte fuerint extremitates longitudinis et latitudinis, unde proposito visui corpore convexitatis modice et quantitatis magne, non planum discernitur eius convexitatis, licet ymago ipsius sit convexa, cum non appareant termini corporis in longitudine vel latitudine. 
◉ Therefore in these [sorts of] mirrors straight lines appear curved, and likewise curved lines appear curved. Moreover, if a curved object is placed before the eye in these mirrors, and if it is long and has some slight breadth, the curvature of that object will appear clearly, since it can be detected by those features lying on or within the body. In fact, unless it is considerable, the curvature is not clearly detected when the boundaries of the length or breadth are hidden [so that the image extends beyond the visible face of the mirror], so when an object of slight curvature and considerable size is placed before the eye, its curvature is not clearly detected, even though its image is curved, since the boundaries along the length and breadth of the object do not appear [in the mirror]. 
◉ Amplius, errores in speculis planis accidentes omnes accidunt et in hiis, et preter illos accidit ymagines linearum rectarum esse curvas, quod a speculis planis est remotum. 
◉ Moreover, all of the errors that occur in plane mirrors occur in these mirrors as well,⁑ and besides those [errors] it happens that the images of straight lines are curved, which is far from the case in plane mirrors. 
◉[Pars quinta] 
◉[Chapter 5 
In speculis columpnaribus exterioribus. 

◉ Amplius, in speculis columpnaribus exterioribus errores accidunt idem qui in speculis spericis exterioribus, linee enim recte videntur curve et diminuta apparet rei vise quantitas, ut in hiis, sed longe fortius quam in eis, quoniam in spericis res magna apparebit quidem minor, sed non multum parva, sed in hiis res etiam maxima videbitur minima. Similiter, linea recta apparebit curva in spericis speculis, sed si modice curvitatis in columpnaribus maxime, unde multiplicantur errores columpnaris speculi super errores sperici. 
◉ Now the same errors occur in convex cylindrical mirrors as occur in convex spherical mirrors, for [in the former] as in the latter, straight lines appear curved and the size of the visible object appears diminished, but far more pronouncedly because in [convex] spherical mirrors a large object will appear smaller, to be sure, but not very small, whereas in convex cylindrical ones even a very large object will appear greatly diminished [in size]. Likewise, a straight line will appear curved in [convex] spherical mirrors, but if it is [even] slightly curved [it will appear] extremely so in [convex] cylindrical [mirrors], so the errors [that occur] in a [convex] spherical mirror are compounded in a [convex] cylindrical mirror. 
◉ Verum in columpnaribus aliquando fit reflexio a linea recta, scilicet a longitudine speculi, aliquando a circulo, aliquando a sectione. Quando linea visa fuerit equidistans longitudini speculi, fiet reflexio a linea longitudinis, et linea visa apparebit recta modice curvitatis. Et hec quidem probabuntur, ad quorum probationem necesse est quiddam premitti, quod hoc est: 
◉ However, in [convex] cylindrical [mirrors] reflection sometimes occurs from a straight line, i.e., [when it occurs] from [a line of] longitude on the mirror, sometimes from a circle [on the mirror’s surface], and sometimes from a [cylindric] section [i.e., an ellipse, on that surface]. When the visible line is parallel to a [line of] longitude on the mirror, the reflection will occur from [that] line of longitude, and a straight visible line will appear [only] slightly curved. These things, moreover, will be demonstrated, but for that demonstration a preliminary point must be made, as follows. 
◉ [PROPOSITIO 16] Sumpta columpnari sectione, et sumpto in ea puncto qui non sit punctus reflexionis, si ab illo puncto ducatur linea ad perpendicularem que est a puncto reflexionis ad axem—et linea illa faciat angulum acutum cum perpendiculari—si ducatur a puncto sumpto linea que sit ortogonalis super contingentem illius puncti, hec linea concurret cum perpendiculari sub axe et sub concursu prioris linee cum perpendiculari. 
◉ [PROPOSITION 16, LEMMA 5] If a cylindric section [e.g., elliptical section APEBR in figure 6.5.16, p. 124] is assumed and some point [E] is taken on it that is not a point of reflection, then, when a line [ED] is extended from that point to the normal [BD dropped] from the point of reflection [B] to the axis [of the cylinder on which the ellipse is chosen]—and that line should form an acute angle [EDB] with the normal—if a line [EU] is drawn perpendicular to the tangent [QEL] at that point [E], this line will intersect the normal [BD] outside the axis and outside the intersection of the previous line [ED] with the normal [BD].⁑ 
◉ Verbi gratia sit AEB [FIGURE 6.5.16, p. 312] sectio, E punctus datus, N punctus visus, B punctus reflexionis, BD perpendicularis, EDB angulus acutus, QEL contingens. 
◉ For example, let AEB be the [assumed cylindric] section, E the given point [that is not the point of reflection], N the visible [object]point, B the point of reflection, BD the [given] normal [dropped from the point of reflection to the axis], EDB an acute angle, and QEL the tangent [to the cylinder as well as to the elliptical section at the chosen point E]. 
◉ Supra B fiat circulus columpne equidistans basi, scilicet BTO, et ducatur a puncto E linea longitudinis columpne, scilicet ET. Ducatur axis DH, et ducatur linea DC perpendicularis supra BD. 
◉ At point B form a circle, i.e., BTO, parallel to the cylinder’s base, and draw a line of longitude through point E on the cylinder, i.e., ET. Draw axis DH [of the cylinder], and draw line DC perpendicular to BD. 
◉ Palam quod superficies HDC est ortogonalis super superficiem circuli. Superficies vero contingens columpnam in puncto B erit equidistans huic superficiei, quoniam linea longitudinis ducta a puncto B erit equidistans axi, et contingens supra B erit equidistans CD. Igitur superficies in qua sunt linee LE, ET non est equidistans superficiei HDC. Igitur concurret cum ea. Concurrat in linea LC, et ducatur linea TC, que quidem erit contingens, cum superficies LET sit contingens. Ducta autem linea TD, erit angulus CTD rectus, quoniam TD dyameter. 
◉ It is obvious that plane HDC is orthogonal to the plane of the circle [BTO]. But the plane tangent to the cylinder at point B will be parallel to this plane [HDC] because the line of longitude extended from point B will be parallel to the axis, and the tangent at [point] B [along the line of longitude dropped through it] will be parallel to CD. Therefore, the plane containing lines LE and ET is not parallel to plane HDC. It will therefore intersect that [plane HDC]. Let it intersect along line LC, and draw line TC, which will of course be tangent [to the cylinder], since plane LET is tangent [to it, by construction]. Moreover, when line TD is drawn, angle CTD will be right because TD is a [a radial segment of the] diameter [of circle BTO, and CT is tangent to the circle at its endpoint]. 
◉ Fiat autem supra E circulus columpne equidistans basi, scilicet ESP. Punctus axis in hoc circulo sit K, et ducatur linea KE. Ducatur etiam linea DL, que quidem secabit superficiem circuli ESP. Secet in puncto F, ubicumque sit punctus extra circulum vel intra, et ducantur linee KF, EF. Et a puncto F ducatur perpendicularis super superficiem circuli BTO que sit FM, et ducatur linea TM. 
◉ Now at point E form a circle, i.e., ESP, on the cylinder parallel to the base. Let K be the point on the [cylinder’s] axis [where it intersects] this circle, and draw line KE. Draw line DL, as well, and it will certainly intersect the plane of circle ESP. Let it intersect at point F, wherever that point may lie either outside or inside the circle, and draw lines KF and EF.⁑ Then from point F draw FM perpendicular to the plane of circle BTO, and draw line TM. 
◉ Palam quoniam KD equidistans et equalis FM, et ita KF equidistans et equalis DM. Similiter KD equidistans et equalis ET, et KE equidistans et equalis DT. Erit ergo TE equidistans et equalis FM, et ita EF equidistans et equalis TM. 
◉ It is evident that KD is parallel and equal to FM [since they are perpendicular to parallel planes], and so KF is parallel and equal to DM. Likewise, KD is parallel and equal to ET, and KE is parallel and equal to DT. Hence, TE will be parallel and equal to FM, and so EF [will be] parallel and equal to TM. 
◉ Verum superficies KDL est ortogonalis super superficiem sectionis BEO, et est ortogonalis super superficiem circuli ESP. Ergo est ortogonalis super lineam communem sectioni et circulo que est EF. Igitur angulus EFK rectus. Similiter angulus TMD rectus. 
◉ But plane KDL is perpendicular to plane BEO of the [cylindric] section, and it is perpendicular to the plane of circle ESP. Therefore, it is perpendicular to common section EF of the [cylindric] section and the circle. So angle EFK is right. Likewise, angle TMD is right [since DM and KF are parallel, as are TM and EF]. 
◉ Cum ergo angulus DTC sit rectus, multiplicatio DM in MC sicut TM in FE, sed quoniam FM equidistans CL, erit proportio DF ad FL sicut DM ad MC. Sed DF maior DM; igitur FL maior MC. Igitur maior est multiplicatio DF in FL quam DM in MC, quare, cum TM sit equalis EF, erit multiplicatio DF in FL maior ductu linee EF in FE, quare angulus LED maior recto, si enim esset rectus, cum linea EF sit perpendicularis super LD, esset ductus DF in FL equalis quadrato EF. Restat ergo ut angulus DEQ sit acutus. Ergo ortogonalis ducta a puncto E, ortogonalis in quam super contingentem QL, cadet sub lineam ED et concurret cum perpendiculari BD sub puncto D, quod est propositum. 
◉ Therefore, since angle DTC is right [by construction], rectangle DM,MC = rectangle TM,FE,⁑ but since FM is parallel to CL [because CL is necessarily parallel to TE, given that it is the common section of planes TCLE and HDT, which are both perpendicular to the plane of circle ESP], then [triangles DFM and DLC will be similar and will have corresponding sides proportional (by Euclid, VI.4), from which it follows that] DF:FL = DM:MC. But DF > DM, so FL > MC. Consequently, rectangle DF,FL > rectangle DM,MC, so, since TM = EF, rectangle DF,FL > rectangle EF,FE [which = rectangle TM,FE, which = rectangle DM,MC], so angle LED > a right angle, for if it were a right angle, then because line EF is perpendicular to LD, rectangle DF,FL would be equal to EF2.⁑ It therefore follows that angle DEQ [adjacent to angle LED] is acute. Hence, the perpendicular [EU] dropped from point E, that perpendicular being perpendicular to tangent QL, will fall outside line ED and will intersect normal BD outside point D, which is what was set out [to be proven].⁑ 
◉ Hiis premissis, accedendum est ad propositum. 
◉ Now that these things have been set out, it is time to get to the proposition. 
◉ [PROPOSITIO 17] Proponatur columpna [FIGURE 6.5.17, p. 313]; linea equidistans axi sit TH. Erit quidem TH equidistans linee longitudinis columpne. 
◉ [PROPOSITION 17] Assume a cylinder [in figure 6.5.17, p. 126], and let TH be a [visible] line parallel to the [cylinder’s] axis [ZK]. TH will of course be parallel to the line of longitude [AG in the same plane with TH and the axis] of the cylinder. 
◉ Si ergo visus fuerit in eadem superficie cum axe et linea TH, poterit quidem reflecti linea, et erit reflexio a linea longitudinis columpne, que linea est communis superficiei in qua sunt visus et axis et superficiei columpne, sicut ostensum est in libro quinto. Sic videbitur linea TH linea recta, quoniam quelibet perpendicularis ducta a puncto linee TH erit in eadem superficie cum visu et axe, et probabitur ymaginem linee TH esse rectam, sicut probatur in speculis planis de visis lineis. 
◉ Therefore, if the center of sight [E] lies in the same plane as the axis and line TH, the [form of the] line can be reflected, and the reflection will occur from the line of longitude on the cylinder, which is the common section of the plane containing the center of sight and the axis and the surface of the cylinder, as was shown in [proposition 28 of] book 5. Line TH will thus appear as a straight line [T’H’] because any normal dropped from a point on line TH [such as TT’ or HH’] will lie in the same plane as the center of sight and the axis, and it will be proven that the image of line TH is straight, just as it is proven for [straight] lines seen in plane mirrors. 
◉ Sit autem visus extra superficiem linee TH et axis, et TH equidistans axi, qui axis sit ZK. Fiat superficies per visum transiens secans superficiem columpne equidistantem basi. Secabit quidem super circulum. Sit circulus ille BF. Aliquis punctus linee HT refertur ad visum ab aliquo puncto huius circuli. Sit a puncto B, et visus sit E. 
◉ Let the center of sight [E in figure 6.5.17a, p, 126] lie outside the plane containing line TH and the axis, and [let] TH be parallel to the axis, which is ZK. Project a plane that passes through the center of sight and cuts the cylinder’s surface parallel to the base. It will of course cut a circle [on that surface]. Let that circle be BF. [The form of] some point on line HT is reflected to the center of sight from some point on this circle. Let it be [reflected] from point B, and let E be the center of sight. 
◉ Punctus ille linee TH sit Q, et ducantur linee EB, QB, et ducatur a puncto B linea longitudinis, que sit ABG, et ducatur a puncto B perpendicularis cadens super axem in puncto L, que sit ML. Et ducatur a puncto E linea equidistans ML, que sit EO, et ducatur QB usque dum concurrat. Sit concursus in puncto O. 
◉ Let Q be the point on line TH [whose form is reflected to E from B], draw lines EB and QB, draw line of longitude ABG from point B, and draw the normal ML through point B that intersects the axis at point L. Then from point E extend line EO parallel to ML, and extend QB until it intersects [EO]. Let the intersection be at point O. 
◉ Palam quoniam angulus QBM equalis est angulo EBM, sed angulus QBM equalis angulo BOE, quia LM equidistans OE. Similiter angulus MBE equalis angulo BEO, quia coalternus. Igitur angulus BOE equalis est angulo BEO, quare BO, BE equalia. 
◉ It is obvious that angle QBM = angle EBM [by construction], but angle QBM = [alternate] angle BOE because LM is parallel to OE [by construction]. Likewise, angle MBE = angle BEO, since [it is] alternate [given the parallelism of ML and EO]. Therefore, angle BOE = angle BEO, so BO and BE are equal [in isosceles triangle BEO]. 
◉ Sumatur autem alius punctus in linea TH, qui punctus sit T, et ducatur linea TO. Palam quoniam linea TH equidistans linee longitudinis, que est AG. Ergo sunt in eadem superficie, et in illa superficie est linea QBO, quare in eadem erit linea TO. Secabit ergo lineam AG. Secet in puncto G, et ducatur linea EG. 
◉ Now choose another point on line TH, let it be point T, and draw line TO. It is clear that line TH is parallel to line of longitude AG [by construction]. Therefore, they lie in the same plane, and line QBO lies in that plane, so line TO will lie in [that] same [plane]. Hence it will intersect line AG. Let it intersect at point G, and draw line EG. 
◉ Palam etiam quoniam linea AG est perpendicularis super superficiem circuli BF sicut axis cui equidistat, et superficies illius superficies EOBF secans scilicet columpnam equidistantem basi. Igitur angulus GBO rectus, et angulus GBE rectus. Ergo quadratum linee GO valet quadratum linee GB et quadratum linee BO. Similiter quadratum GE valet quadrata GB et BE, et quoniam BE, BO equalia et GB communis, erit GO equalis GE. Igitur angulus GOE equalis angulo GEO. 
◉ It is also clear that line AG is perpendicular to the plane of circle BF, as is the axis [ZK] to which it is parallel, and its plane [is] EOBF [which] cuts the cylinder parallel to its base. Thus, angle GBO [is] right, and [so] angle GBE is right. Consequently GO2 = GB2 + BO2 [by Euclid, I.47]. Likewise, GE2 = GB2 + BE2, and since BE and BO are equal [by previous conclusions], while GB is common, GO = GE. Hence, angle GOE = angle GEO [in isosceles triangle GEO]. 
◉ Ducta autem perpendiculari ZGN, erit equidistans EO, cum sit equidistans MBL. Igitur angulus TGN equalis angulo GOE, et angulus NGE equalis angulo GEO, quare angulus TGN equalis angulo NGE. Cum autem E, O, N, G, Z sint in eadem superficie, et in illa sit G, E, G, T erunt in eadem superficie, et ita in eadem superficie sunt linee EG, NG, TG. Igitur T refertur ad E a puncto G. 
◉ Moreover, if normal ZGN is drawn, it will be parallel to EO, since it is parallel to MBL [to which EO was made parallel by construction]. Therefore, angle TGN = [alternate] angle GOE, and angle NGE = [alternate] angle GEO, so angle TGN = angle NGE. Furthermore, since E, O, N, G, and Z lie in the same plane, and since G lies in that plane, E, G, and T will lie in the same plane, and so lines EG, NG, and TG lie in the same plane [which is therefore the plane of reflection]. Thus, [the form of] T is reflected to E from point G. 
◉ Sumpto autem in linea TH puncto H eiusdem longitudinis a puncto Q cuius est punctus T, et ducta linea HO, transibit quidem per punctum linee AG. Transeat per punctum A. Ducta perpendiculari DA et lineis EA, HAO, erit sicut prius probare quod duo anguli ABO, ABE recti, et duo latera AO, AE equalia, et duo anguli HAZ EAZ equales. Et ita H refertur ad E a puncto A. Similiter sumpto quocumque puncto linee TH, erit probare quod refertur ad E ab alio puncto linee AG, quare linea TH refertur a linea longitudinis que est AG. 
◉ Furthermore, if point H is taken on line TH at the same distance from point Q as point T, and if line HO is drawn, it will pass through [some] point on line AG. Let it pass through point A. When normal DA[Z’] and lines EA and HAO are drawn, it will be a matter of proving as before that the two angles ABO and ABE are right, that the two sides AO and AE are equal, and that the two angles HAZ’ and EAZ’ are equal. And so [the form of point] H is reflected to [point] E from point A. Likewise, if any [other] point on line TH is chosen, it will be a matter of proving that [its form] is reflected to E from another point on line AG, so [the whole form of] line TH is reflected from line of longitude AG. 
◉ [PROPOSITIO 18] Restat probare ymaginem linee TH esse curvam. Palam ex predictis quoniam Q refertur ad E a puncto B, qui est punctus circuli. Sed cum sic refertur a circulo, si ducatur linea a puncto Q ad centrum illius circuli, concurret cum perpendiculari ducta a puncto B, et erit concursus in puncto axis. Ducatur ergo QL concurrens cum ML in puncto axis qui est L, et est centrum circuli FB. Et producatur EB usque concurrat cum QL. Sit concursus in puncto C. Erit C ymago Q, et est C in superficie in qua sunt linee QH, et axis, et linea longitudinis AG. 
◉ [PROPOSITION 18] It remains to demonstrate that the image of line TH is curved. It is clear from the preceding [theorem] that [the form of] Q [in figure 6.5.18, p. 127] is reflected to E from point B, which is a point on circle [FB]. But since it is reflected in this way from the circle, if a line is drawn from point Q to the center of that circle, it will meet the normal [MBL] dropped from point B, and the intersection [of these two lines] will lie at a point on the axis. So draw QL intersecting ML at point L on the axis, and [this] is the center of circle FB. Then extend EB until it meets QL. Let the intersection be at point C. C will be the image of Q, and C lies in the same plane with lines QH and the axis [ZK], and [with] line of longitude AG.⁑ 
◉ Palam etiam quod T refertur ad E a puncto sectionis columpne, scilicet a puncto G. Est autem a puncto T lineam ducere perpendicularem super lineam contingentem in puncto alio sectionis, que quidem concurret cum perpendiculari ducta a puncto G, que est NGZ, sub axe, id est sub puncto Z, qui est concursus perpendicularis NZ et axis, quoniam ducta linea TZ, erit angulus TZN acutus. Ducatur ergo TX concurrens cum NZ in puncto X, et producatur EG donec concurrat cum TX in puncto I. Erit I ymago puncti T. 
◉ It is also evident that [the form of] T is reflected to E from a point on a [cylindric] section of the cylinder, namely, point G. Moreover, [as established in proposition 16, lemma 5] a line can be drawn from point T perpendicular to a line tangent to another point on the [cylindric] section, and it will intersect normal NGZ dropped from point G outside the axis, that is, outside point Z, which is the intersection of normal NZ and the axis, for if line TZ is drawn, angle TZN will be acute [as stipulated in proposition 16]. Accordingly, draw TX [normal to the cylindric section, as prescribed, and] intersecting NZ at point X, and extend EG until it intersects TX at point I. I will be the image of point T. 
◉ Similiter ducta a puncto H linea, que sit ortogonalis super punctum sectionis a qua refertur, concurret cum perpendiculari DAZ sub puncto D, que est punctus axis. Concurrat in puncto P, et producatur EA donec concurrat cum HP in puncto S. Erit ymago puncti H punctus S. Ducatur autem linea SI. 
◉ Likewise, when the line [HP] orthogonal at a point on the [cylindric] section from which reflection [of the form of point H] occurs is drawn from H, it will intersect normal DAZ’ outside of point D, which is a point on the axis. Let it intersect at point P, and extend EA until it intersects HP at point S. Point S will be the image of point H. Now draw line SI. 
◉ Palam cum linea TI concurrat cum perpendiculari NZ, que est equidistans linee EO, concurret cum linea EO. Similiter linea HS; quoniam concurrit cum perpendiculari DAZ, que est equidistans EO, concurret cum EO. Sed quoniam situs T respectu puncti E idem est cum situ H et eadem longitudo, similiter situs puncti T et puncti H ad punctum O idem, et punctorum I, S respectu O etiam est idem. Erit idem situs linearum TI, HS respectu linee EO. 
◉ It is clear that, since line TI intersects normal NZ, which is parallel to line EO, it will intersect line EO. The same holds for line HS; because it intersects normal DAZ, which is parallel to EO, it will intersect EO. But since T’s location with respect to point E is equivalent to and the same distance [from E] as H’s location [by construction], the location of point T and of point H [will] likewise [be] equivalent with respect to point O, and [that of] points I and S is also equivalent with respect to O. The location of lines TI and HS will also be equivalent with respect to line EO.⁑ 
◉ Igitur linee TI, HS concurrent super idem punctum linee EO. Concurrant in puncto U. Erit TUH triangulus, et in superficie huius trianguli erit linea IS. Axis autem non est in hac superficie. 
◉ Lines TI and HS will therefore intersect at the same point on line EO [since each lies in a plane with it, and both are inclined toward one another]. Let them intersect at point U. TUH will [therefore] be a triangle, and [straight] line IS will lie in the plane of this triangle. The axis, however, does not lie in this plane [since normals TIU and HSU bypass it]. 
◉ Verum TH est in eadem superficie cum axe; ergo superficies illa secat superficiem trianguli super lineam communem que est TH, non super aliam. Cum ergo punctus C sit in superficie linee TH et axis, et non sit in linea TH, non est in superficie trianguli TUH, et duo puncta I, S sunt in superficie illius trianguli, quare linea ICS est curva, et ymago linee TH erit curva, quod est propositum. 
◉ But TH does lie in the same plane as the axis so that plane [TZKH] intersects the plane of the triangle [TUH] along common section TH, not along some other [line of section]. Therefore, since point C lies in the plane of line TH and the axis, but not on line TH [itself], it does not lie in the plane of triangle TUH, whereas the two points I and S do lie in the plane of that triangle, so line ICS is curved, and the image of line TH will [therefore] be curved, which is what was set out [to be proven]. 
◉ Sed eius curvitas est modica, quia perpendicularis ducta a puncto C ad superficiem circuli est valde parva, et quanto maior fuerit linea visa equidistans linee longitudinis speculi, tanto ymago eius erit minus curva, quanto minor magis. 
◉ But its curvature is slight because the perpendicular dropped from point C to the plane of the circle is extremely small,⁑ and the closer the visible line is to being parallel to the line of longitude on the mirror, the less sharply curved it[s image] is, [whereas] the farther [it is from such parallelism] the more [sharply curved its image is].⁑ 
◉ [PROPOSITIO 19] Amplius, si linea TH [FIGURE 6.5.19, p. 314] secet superficiem in qua sunt centrum visus et axis, et sit ortogonalis super eam, visus aut erit in illa superficie linee TH secante ortogonaliter superficiem axis et visus, aut extra. 
◉ [PROPOSITION 19] Furthermore, if line TH intersects the plane containing the center of sight and the axis, and if it is orthogonal to it, the center of sight will either lie in the plane of line TH intersecting the plane of the axis and the center of sight orthogonally, or [it will lie] outside [that plane]. 
◉ Si fuerit in illa superficie, aut erit supra lineam TH, aut infra. Si supra, cum illa linea sit corporalis, occultabit visui speculum, et ita non reflectetur, sed forsan capita eius apparebunt et reflectentur a circulo columpne qui communis est superficiei linee TH secanti columpnam et columpne. Et erit horum capitum ymago sicut in spericis exterioribus. 
◉ If [the center of sight] lies in that plane, it will lie beyond or in front of line TH. If [it lies] beyond, then, since that line has bodily dimensions, it will block the mirror from the center of sight, and so it[s form] will not be reflected [to the eye], although perhaps its terminal segments will appear and be reflected from the circle on the cylinder that forms the common section of the plane of line TH that cuts the cylinder and the cylinder [itself]. And the image of these terminal segments will [appear] just as [they do] in convex spherical [mirrors, as described in proposition 14, paragraph 4.138].⁑ 
◉ Similiter, si visus fuerit sub linea TH, occultabitur pars eius propter capud in quo est visus. Pars autem linee visa refertur a circulo eodem penitus modo quo in exterioribus spericis. 
◉ Likewise, if the center of sight lies in front of TH, part of that line will be hidden by the head containing the center of sight. Nonetheless, the visible part of the line is reflected [to the center of sight] from the circle [formed by the plane of reflection] in exactly the same way as in convex spherical [mirrors, according to proposition 14, paragraph 4.138]. 
◉ Si vero visus fuerit extra superficiem linee TH ortogonaliter secantem superficiem visus et axis, sit E visus, et XZG columpna. Refertur H ad E ab aliquo puncto columpne. Sit a B. Sit T eiusdem longitudinis a puncto E. Dico quod T refertur ad E ab alio puncto columpne, et cum puncta H, T sint eiusdem situs et eiusdem longitudinis a puncto E, erunt similiter puncta reflexionum, scilicet B, G, eiusdem longitudinis et eiusdem situs a puncto E. Igitur duo puncta B, G erunt in circulo. 
◉ But if the center of sight lies outside the plane of TH that cuts the plane of the center of sight and the axis orthogonally, then let E [in figure 6.5.19, p. 128] be the center of sight [above line TH] and XZG the cylinder.⁑ [The form of point] H is reflected to E from some point on the cylinder. Let [it be reflected] from B. Let T lie the same distance [as H] from point E. I say that [the form of point] T is reflected to E from another point [i.e., other than B] on the cylinder, and that, since points H and T are equivalently situated and the same distance from point E, their points of reflection, i.e., B and G, will be equivalently situated and the same distance from point E. Therefore, the two points B and G will lie on a circle. 
◉ Sit circulus BZG, eius centrum D. Ducantur linee HB, BE, TG, GE, et a centro ducantur perpendiculares supra contingentes B, G, scilicet DBO, DGS. Et ducatur linea ED, et producantur HB, TG usque concurrant cum linea ED. 
◉ Let BZG be the circle, with D its centerpoint. Draw lines HB, BE, TG, and GE, and from the centerpoint [D] draw normals to the tangents at B and G, i.e., [normals] DBO and DGS. Then draw line ED, and extend HB and TG until they intersect line ED. 
◉ Cum puncta H, T sint eiusdem situs et longitudinis respectu E et respectu D, et similiter puncta B, G eiusdem situs respectu D et respectu E, habebunt linee HB, TG eundem situm respectu linee ED, et ita concurrent in idem punctum illius linee. Sit in puncto L. 
◉ Since points H and T are equivalently situated and the same distance with respect to E and with respect to D, and since, by the same token, points B and G are equivalently situated with respect to D and with respect to E, lines HB and TG will be equivalently situated with respect to line ED, and so they will intersect at the same point on that line. Let [that intersection] be at point L. 
◉ Fiat linea longitudinis columpne in qua punctus Z, et sit hec linea in superficie visus et axis, que sit AZ, et ducantur LZN, DZC. Q sit punctus linee TH, punctus scilicet qui est in superficie visus et axis, et a puncto Q ducatur equidistans linee DZC. Cadet quidem hec linea super axem, et LZN cadet in hanc lineam supra punctum Q. Cadat in puncto N. 
◉ Produce the line of longitude on the cylinder that contains point Z, let this line lie in the plane containing the center of sight and the axis, let it be AZ, and draw [lines] LZN and DZC. Let Q be a point on line TH, that is, the point [on it] that lies in the plane of the center of sight and the axis, and from point Q draw a line parallel to line DZC. This line will fall on the axis, and LZN will fall on this line beyond point Q. Let it fall at point N. 
◉ Palam ex predictis quod angulus HBO equalis angulo OBE. Sed angulus HBO equalis angulo LBD per contrapositionem, et angulus OBE equalis duobus angulis BED, BDE, quia extrinsecus. Ergo angulus LBD equalis duobus angulis BED, BDE. Fiat ergo angulus MBD equalis angulo BDE. Remanet angulus MBL equalis angulo BEL, quare ductus EM in ML equalis quadrato BM. 
◉ It is clear from the foregoing that angle [of incidence] HBO = angle [of reflection] OBE [by construction]. But angle HBO = angle LBD because they are vertical [angles], and angle OBE = the two angles BED + BDE because it is external [to triangle BDE and therefore equal to the two opposite interior angles, by Euclid, I.32]. Therefore, angle LBD = the two angles BED + BDE. So form angle MBD = angle BDE. It follows that angle MBL = angle BEL [i.e., BED], so the rectangle EM,ML = BM2.⁑ 
◉ Ducatur linea MZ. Quoniam angulus BDM maior angulo ZDM, et duo latera ZD, DM equalia duobus lateribus BD, DM, erit MB maior MZ, quare ductus EM in ML maior quadrato MZ. Sit ductus EM in MI equalis quadrato MZ, et ducantur linee IB, IZ. Erit ergo angulus MZI equalis angulo ZEI, quare MZL maior angulo ZED. 
◉ Draw line MZ. Since angle BDM > angle ZDM, and since the two sides ZD and DM [of triangle ZDM] are equal, respectively, to the two sides BD and DM [of triangle BDM], MB > MZ, so the rectangle EM,ML > MZ2.⁑ Let the rectangle EM,MI = MZ2 [which means that EM:MZ = MZ:MI] and draw lines IB and IZ. Hence, angle MZI = angle ZEI [because triangles MZE and MZI are similar according to the proportionality of sides EM and MZ in triangle MZE and sides MZ and MI in triangle MZI] , so [angle] MZL > angle ZED. 
◉ Sed quoniam angulus MBD positus est equalis angulo BDM, erit linea MD equalis linee MB. Sed MB maior MZ, quare MD maior MZ. Igitur angulus MZD maior angulo MDZ; igitur angulus DZL maior duobus angulis ZDE, ZED. Sed angulus DZL equalis angulo NZC, et angulus CZE equalis duobus angulis ZDE, ZED, quare angulus NZC maior angulo CZE. 
◉ But since angle MBD has been posited equal to angle BDM [by construction], line MD = line MB [in isosceles triangle DMB]. But MB > MZ [by previous conclusions], so MD > MZ. Therefore, angle MZD > angle MDZ [since it is subtended by a longer line], so angle DZL > the two angles ZDE + ZED.⁑ But angle DZL = [vertical] angle NZC, and [exterior] angle CZE = the two [interior and opposite] angles ZDE + ZED [in triangle ZED], so angle NZC > angle CZE. 
◉ Secetur ad equalitatem per lineam FZ, que quidem concurret cum linea NQ supra punctum N. Cum ergo angulus FZC equalis angulo CZE, refertur F ad E a puncto Z. Q refertur ad E a puncto linee longitudinis que transit per Z, a puncto que est AZ, scilicet ultra Z. Si enim a puncto citra Z, id est propinquiori E, linea ducta a puncto Q ad punctum illud reflexionis secabit lineam FZ, et ita punctus sectionis refertur ad E a duobus punctis, quod est impossibile. 
◉ Let [angle FCZ] equal [to angle CZE] be cut [from NCZ] by line FZ, which will intersect line NQ [at point F] beyond point N. Therefore, since angle FZC = angle CZE, [the form of] F is reflected to E from point Z. [The form of point] Q is reflected to E from a point on the line of longitude passing through Z, that is, from a point on AZ beyond [i.e., below] Z. For if [it occurs] from a point this side of Z, i.e., nearer E, the line extended from point Q to that point of reflection will cut line FZ, and so the point of intersection is reflected to E from two points, which is impossible.⁑ 
◉ Sumatur ergo ultra Z punctus K a quo refertur Q ad E, et ducatur linea EK donec concurrat cum linea NQ in puncto P. Erit P ymago Q. Sed H refertur ad E a puncto sectionis columpne. Si ergo a puncto H ducatur perpendicularis super contingentem sectionem in aliquo puncto, perpendicularis illa concurret cum perpendiculari CZD sub axe. Concurrat in puncto U. 
◉ Take point K below Z from which [the form of] Q is reflected to E, then, and extend line EK until it intersects line NQ [i.e., the cathetus dropped from objectpoint Q] at point P. P will be the image of Q. But [the form of] H is reflected to E from a point on the cylindric section [formed on the cylinder by plane of reflection HBE]. Therefore, if a normal is dropped from point H to the line tangent to the [cylindric] section at some point [on it], that normal [i.e., the cathetus] will intersect normal CZD outside the axis [by proposition 16, lemma 5]. Let it intersect at point U. 
◉ Similiter a puncto T est ducere unam perpendicularem super sectionem a cuius puncto refertur ad E. Et quoniam puncta H, T sunt eiusdem situs respectu linee CZD, et puncta sectionis similiter per que transeunt perpendiculares, igitur ille due perpendiculares concurrent in idem punctum linee CZD. Concurrant ergo in puncto U. 
◉ Likewise, from point T a normal can be drawn to the [cylindric] section from a point on which [its form] is reflected to E. And since points H and T are equivalently situated with respect to line CZD, the same also holds for the points on the [cylindric] section through which the normals [i.e., the catheti] pass, so those two normals will intersect at the same point on line CZD. Accordingly, let them intersect at point U. 
◉ Linea EB concurret cum linea HU. Sit concursus in puncto R. Similiter EG concurrat cum TU in puncto Y, et ducatur linea RY. Palam quod R est ymago H, Y est ymago T, et habemus triangulum ERY. Extra superficiem huius trianguli est punctum Z, et ita superficies huius trianguli altior est linea EP, et ita P extra. Quare linea RPY erit curva, et illa est ymago linee TH, et est quidem hec ymago curvitatis non modice, quod est propositum. 
◉ [Therefore, the extension of] line EB will intersect line HU. Let R be the point of intersection. By the same token, let EG intersect TU at point Y, and draw line RY.⁑ It is obvious that R is the image of H, [and] Y is the image of T, and we have triangle ERY. Point Z lies outside the plane of this triangle, so the plane of this triangle is higher than line EP, and so P lies outside [that plane]. Hence, line RPY will be curved, and it is the image of line TH, and the curvature of this image is certainly not inconsiderable, which is what was set out [to be proven]. 
◉ Palam ergo quod in hiis speculis, si linea recta visa equidistans fuerit linee longitudinis columpne, erit ymago eius aut recta aut accedens ad rectitudinem. Si vero linea visa recta equidistans fuerit latitudini columpne, erit ymago eius curva curvitate non modica. 
◉ It is therefore clear that in these [sorts of] mirrors, if a straight visible line is parallel to a line of longitude on the cylinder, its image will be either straight or verging toward straightness. But if a straight visible line is parallel to the width of the mirror [i.e., the plane through it is perpendicular to the cylinder’s axis], its image will be curved, and its curvature will not be inconsiderable. 
◉ Linee autem inter has duas site, que magis accedunt ad situm linee equidistans longitudini columpne erunt ymagines earum rectitudini magis vicine, et ymagines earum que propinquiores sunt situi equidistantium latitudini erunt magis curve. Et minuetur vel augmentabitur curvitas ymaginum secundum accessum vel elongationem linearum ad alterum horum situum, et hoc est propositum. 
◉ Furthermore, among [visible] lines oriented between these two [extremes], the images of those that verge more closely toward an orientation parallel to the longitude of the cylinder will be closer to straight, whereas the images of those that are nearer to an orientation parallel to the [cylinder’s] width will be more curved. And the curvature of the images will diminish or augment depending on how close or far the lines are from either of these orientations, and this is what was set out [to be demonstrated]. 
◉[Pars sexta] 
◉[Chapter 6] 
In speculis piramidalibus exterioribus. 

◉ Amplius in speculis piramidalibus exterioribus idem errores accidunt qui in columpnaribus exterioribus eveniunt, linee enim vise equidistantes longitudinis piramidis aut recte videntur, aut forte equidistantes latitudini curve, et intermedie augmentant vel diminuunt curvitatem secundum propinquitatem harum vel harum remotionem, et hoc quidem probabitur. Quoddam tamen premittendum proponamus et est: 
◉ Moreover, in convex conical mirrors the same errors occur as happen in convex cylindrical [mirrors],⁑ for [straight] visible lines that are parallel to the longitude of the cone appear straight, whereas those parallel to the width [of the cone appear] curved, and for those at intermediate positions, their curvature augments or diminishes according to how near or how far [they are from those extreme positions], and this will of course be demonstrated. However, we must set forth something beforehand, and it is [as follows]. 
◉ [PROPOSITIO 20] Si sumatur in superficie piramidis punctus reflexionis et fiat sectio transiens per punctum illud, et in sectione sumatur punctus remotior ab acumine piramidis puncto reflexionis et a puncto sumpto ducatur perpendicularis super contingentem sectionem, hec perpendicularis concurret cum perpendiculari ducta a puncto reflexionis sub axe. 
◉ [PROPOSITION 20, LEMMA 6] If a point of reflection is taken on the surface of a cone and a [conic] section is produced [on that surface] to pass through that point, and if a point is taken on that [conic] section farther from the vertex of the cone than the point of reflection and a normal is dropped from the selected point to a line tangent to the [conic] section, this normal will intersect the normal dropped from the point of reflection [at a point] outside the axis. 
◉ Verbi gratia, sit ABGZ [FIGURE 6.6.20, p. 315] piramis erecta super bases suas, A acumen piramidis, BFZ sectio, E punctus reflexionis, Z punctus sectionis remotior a puncto A quam E. Supra punctum Z sit superficies secans piramidem equidistans basi. Secabit quidem supra circulum communem. Sit circulus ille GBRZ, et ducantur linee AZ, AE, et producatur AE donec sit equalis AZ. Veniet quidem ad circulum. Cadat ergo in puncto eius O. 
◉ For instance, let ABGZ [in figure 6.6.20, p. 129] be a cone standing upright on its bases [i.e., a right cone], A the cone’s vertex, BFZ the [conic] section [produced on its surface], E the point of reflection, and Z the point on the [conic] section farther from [vertex]point A than E. At point Z let there be a plane cutting the cone parallel to its base. It will of course cut it along a circle [forming the] common [section of the cutting plane and the cone’s surface]. Let that circle be GBRZ, draw lines AZ and AE, and extend AE until it is equal to AZ. It will reach the circle. So let it fall at point O on it. 
◉ C sit centrum circuli, et ducatur axis AC, et a puncto E ducatur perpendicularis super superficiem contingentem piramidem. Concurret quidem cum axe circa centrum circuli quod est C. Sit in puncto D, et ducatur linea DZ. 
◉ Let C be the center of the circle, draw axis AC, and from point E draw the normal [ED] to the plane tangent to the cone [at that point]. It will of course intersect the axis in the vicinity of the circle’s centerpoint C. Let [that intersection] be at point D, and draw line DZ.⁑ 
◉ Et a puncto O ducatur perpendicularis concurrens cum axe in puncto K, et ducantur linee DZ, KZ. Et supra punctum Z ducatur contingens sectionem, que sit TQ, et alia contingens circulum BGZ, que sit ZY. 
◉ Then from point O draw a normal [OK] intersecting the axis at point K, and draw lines DZ and KZ. At point Z produce [line] TQ tangent to the [conic] section, and [at the same point produce] another [line] ZY tangent to circle BGZ. 
◉ Et ducatur linea BCZ, et a puncto C ducatur perpendicularis super lineam BCZ, que sit CR. Erit quidem perpendicularis super axem, cum axis sit perpendicularis super superficiem circuli, quare CR est perpendicularis super superficiem ACZ. Et erit equidistans contingenti ZY, quare ZY est perpendicularis super superficiem ACZ, quare TQ non est perpendicularis super eandem superficiem. 
◉ Next draw line BCZ, and from point C draw CR perpendicular to line BCZ. It will of course be perpendicular to the axis, since the axis is perpendicular to the circle [in whose] plane [CR lies], so CR is perpendicular to plane ACZ. It will also be parallel to tangent ZY, so ZY is perpendicular to plane ACZ, [and] so TQ is not perpendicular to that same plane. 
◉ Verum quoniam K est polus ad circulum BRZ, cum linee KO, KZ sint equales, et axis AK communis, erit angulus AOK equalis angulo AZK, et ita angulus AZK rectus. Cum ergo linea KZ sit perpendicularis super AZ, que est linea longitudinis, erit perpendicularis super superficiem contingentem piramidem super hanc lineam longitudinis. Sed TQ est in superficie contingenti, quia est communis superficiei contingenti et sectioni. Igitur KZ est perpendicularis super TQ. 
◉ However, because K is the [the endpoint of] pole [KC] in circle BRZ, then, since lines KO and KZ are equal [because they are lines of longitude on a right cone with its vertex at K and its base circle passing through Z and O], and since axis AK is common [to triangles AOK and AZK], angle AOK = angle AZK, and so angle AZK is right [since angle AOK was constructed as a right angle]. Therefore, since line KZ is perpendicular to [line] AZ, which is a line of longitude [on the mirror], it will be perpendicular to the plane tangent to the cone[’s surface] along this line of longitude. But TQ lies in the [same] tangent plane because it is the common [section] of the tangent plane and the [conic] section. Accordingly, KZ is perpendicular to TQ. 
◉ Ducatur autem HZ in superficie sectionis perpendicularis super lineam TQ. Cum autem linea KZ sit extra superficiem sectionis, secabit lineam HZ, nec erit una linea cum illa. Superficies ergo KZH secat superficiem sectionis super lineam communem HZ, et secat lineam TQ super punctum Z. Et superficies AZK secat superficiem AZH super lineam communem KZ. 
◉ Furthermore, draw HZ in the plane of the [conic] section perpendicular to line TQ [and therefore normal to the section itself]. Since line KZ lies outside the plane of the [conic] section, it will intersect line HZ and will [therefore] not form a single line with it. Hence, plane KZH intersects the plane of the [conic] section along common section HZ, and it intersects line TQ at point Z. In addition, plane AZK intersects plane AZH along common section KZ. 
◉ Verum DZ est in superficie sectionis, et secatur a linea KZ in puncto Z, et punctus T supra superficiem KZH, punctus Q infra. Et ita superficies KZH secat superficiem DZQ super lineam communem, et illa linea communis est perpendicularis super lineam TQ, quia linea illa est in superficie HZK super quam est perpendicularis TQ. Et quoniam superficies HZK secat superficiem DZQ, et declinatio superficiei HZK sit ex parte ZE, erit linea communis sectioni illarum superficierum inter lineas QZ, DZ, et ita concurret cum perpendiculari ED sub axe. Et quod necessario concurrat cum ea probatum est in libro quinto, figura 19, et ita propositum. 
◉ But DZ lies in the plane of the [conic] section, and it is intersected by line KZ at point Z, and point T [lies] above plane KZH, point Q below [it, i.e., on the other side of it from T]. And so plane KZH cuts plane DZQ along a common section, and that common section is perpendicular to line TQ because that line lies in plane HZK to which TQ is perpendicular. And since plane HZK intersects plane DZQ, and since plane HZK slants in the direction of [segment] ZE [of the conic section], the common section [HZX] of those planes will lie between lines QZ and DZ, and so it will intersect normal ED [at point X] outside the axis. That it [i.e., the normal to the section at Z] must necessarily intersect it [i.e., normal ED dropped from center of sight E] has been demonstrated in book 5, proposition 26,84 and so what was set out [to be demonstrated has been shown].⁑ 
◉ [PROPOSITIO 21] Sit ergo piramis cuius acumen A [FIGURE 6.6.21, p. 316], axis AH, linea longitudinis AZ, et a puncto Z ducatur perpendicularis supra superficiem contingentem piramidem in linea AZ, que necessario concurret cum axe. Sit linea TZH. 
◉ [PROPOSITION 21] So let there be a [right] cone [in figure 6.6.21, p. 129] with its vertex at A, AH being its axis and AZ a line of longitude, and from point Z to the plane tangent to the cone along line AZ drop a perpendicular, which will perforce intersect the axis [at point H]. Let it be line TZH. 
◉ Ducatur a puncto A linea extra piramidem supra superficiem contingentem piramidem in linea AZ faciens angulum acutum cum axe et cum linea longitudinis AZ, que sit AN. Et in superficie AHN a puncto H ducatur linea cum axe faciens angulum equalem angulo AHZ, que linea necessario concurrat cum linea AN, que sit HO. Et facto supra punctum Z circulo equidistans basi, transibit HO per circulum sicut HZ transit per ipsum. 
◉ From point A extend line AN outside the cone [and] above the plane tangent to the cone along line AZ so that it forms an acute angle with the axis, as well as with line AZ of longitude. From point H within plane AHN, draw line HO forming an angle [AHO] equal to angle AHZ, that line necessarily intersecting line AN [at point O. Consequently] when a circle is produced through point Z parallel to the [cone’s] base, HO will pass through [that] circle just as HZ passes through it. 
◉ Ducatur autem linea OZ, et producatur usque ad punctum F. Quoniam linea OZ est supra superficiem contingentem piramidem in linea AZ, cum linea HZ sit perpendicularis supra illam superficiem, erit angulus OZH maior recto. Igitur angulus FZH acutus. 
◉ Now draw line OZ, and extend it to point F. Since line OZ lies above the plane tangent to the cone along line AZ, then because HZ is perpendicular to that plane [by construction], angle OZH will be greater than a right angle [because it intersects the plane tangent to AZ from above]. Consequently, [adjacent] angle FZH is acute [so that ZF lies inside the cone]. 
◉ A puncto Z ducatur contingens supra circulum, que sit ZM, et a puncto F ducatur perpendicularis supra AZ cadens in puncto eius E, que producta concurret cum AO, quoniam angulus OAZ est acutus. Concurrat ergo in puncto N, et a puncto E ducatur equidistans linee TH, et sit QE. 
◉ From point Z draw ZM tangent to the circle, and from point F draw a line perpendicular to AZ to fall at point E on it, and when it is extended, it will intersect AO because angle OAZ is acute [by construction, and because AO, AZ, and OZF lie in the same plane]. Accordingly, let it fall at point N, and from point E draw line QE parallel to line TH. 
◉ Et a puncto E ducatur equidistans linee MZ, que sit LE. Palam quoniam MZ est perpendicularis supra AE, quoniam est perpendicularis supra TH et supra dyametrum circuli, cuius est contingens. Igitur LE est perpendicularis super AE. 
◉ Then from point E draw LE parallel to line MZ. It is evident that MZ is perpendicular to AE because it is perpendicular to TH, as well as to the diameter of the circle, to which it is tangent. Therefore, LE is perpendicular to AE [since it is parallel to MZ, by construction]. 
◉ Fiat autem superficies LQD secans piramidem. Erit quidem sectio piramidalis. Cum ergo AE sit perpendicularis super FN, et super QD, et super LE, erit FN in superficie illa secante piramidem. Fiat ergo CF equidistans QE. Erit quidem equidistans TZ. 
◉ Now produce plane LQD cutting the cone. It will of course form a conic section [because the cutting plane intersects the axis below the circle passing through E]. Hence, since AE is perpendicular to FN [by construction], as well as to QD and LE, FN will lie in that plane [QEDL], which cuts the cone [along the aforementioned conic section]. Accordingly, produce CF parallel to QE. It will of course be parallel to TZ. 
◉ Verum cum angulus OZT sit acutus, angulus TZF sit obtusus. Ducatur a puncto Z linea faciens cum TZ angulum equalem angulo OZT, que quidem linea necessario secabit FC. Secet in puncto C, et ducatur linea EC. Cum ergo CZ, OZ sint in eadem superficie, et angulus OZT equalis angulo TZC, punctus O refertur ad C a puncto Z. 
◉ But since angle OZT is acute [by previous conclusions, adjacent] angle TZF is obtuse. From point Z draw a line [ZC] that forms with TZ an angle [TZC] equal to angle OZT, and that line will necessarily intersect FC. Let it intersect at point C, and draw line EC. Therefore, since CZ and OZ lie in the same plane, and since angle OZT = angle TZC [by construction, the form of] point O is reflected to C from point Z. 
◉ Verum quoniam angulus OZT equalis angulo ZFC, et angulus OZT equalis angulo ZCF, erunt latera ZC, ZF equalia, et quia angulus FEZ rectus, quadratum FZ valet quadrata EZ,EF, et quadratum CZ valet quadrata EZ, EC. Igitur CE, FE equalia, et ita anguli ECF, EFC equales, quare anguli NEQ QEC equales. Et cum in eadem superficie sint C, E, N, refertur N ad C a puncto E. 
◉ Since, however, angle OZT = [alternate] angle ZFC, and since angle OZT = [alternate] angle ZCF, sides ZC and ZF [of triangle ZCF] will be equal, and given that angle FEZ is right [by construction], FZ2 = EZ2 + EF2 [by Euclid, I.47], while CZ2 = EZ2 + EC2 [by the same theorem]. Therefore, CE and FE are equal, and so angles ECF and EFC are equal, whereby angles NEQ [which is alternate to EFC] and QEC [which is alternate to ECF] are equal. And since C, E, and N lie in the same plane [i.e., the plane producing the conic section through point E, the form of] point N is reflected to C from point E. 
◉ Similiter ducatur a puncto F quecumque linea ad aliquod punctum linee ZE, et producatur usque ad ON. Probabitur de puncto linee ON in quam cadit quod refertur ad C a puncto ZE, quoniam secat illa linea. Simili modo et omnium huiusmodi linearum probatio sumet initium a perpendiculari, que est FE, et a parte linee EZ, que erit terminus, et ita quodlibet punctum linee ON refertur ad C ab aliquo puncto linee EZ. 
◉ Likewise, draw some line from point F to some point on line ZE, and extend it to ON. As to the point on line ON where it falls, it will be proven that [its form] is reflected to C from the [corresponding] point on ZE because it cuts that line. In the same way, as well, for all such lines, the proof will take its start from perpendicular FE with respect to line EZ, which will be the common terminal [baseline], and so [the form of] any given point on line ON is reflected to C from some point on line EZ. 
◉ [PROPOSITIO 22] Hoc ergo declarato, dicamus: cum visus comprehenderit lineas rectas transeuntes per caput speculi piramidalis convexi recti obliquas super axem speculi, in hoc speculo tunc forme earum erunt parum convexe. 
◉ [PROPOSITION 22] Having demonstrated this point, we should state [that], when the eye perceives straight lines that pass through the vertex of a right convex conical mirror at a slant to the mirror’s axis, the forms of those [lines] will be somewhat convex in that mirror. 
◉ Sit ergo speculum piramidale erectum ABG [FIGURE 6.6.22, p. 316], cuius caput sit A, et cuius axis sit AD, et extrahamus in superficie eius lineam AZ, quocumque modo sit, in qua signetur punctum Z, quocumque modo sit. Et transeat per Z superficies equidistans basi piramidis, et faciat circulum ZU. Et extrahamus ex Z perpendicularem ZH super AZ. Hec ergo linea concurret cum axe piramidis, et concurrat ergo in H. 
◉ Accordingly, let the right conical mirror be ABG [in figure 6.6.22, p. 130], with A as its vertex and AD as its axis, let us produce line AZ at random on its surface, and let point Z be marked on it at random. Let a plane pass through Z parallel to the base of the cone, and let it form circle ZU. Then from Z let us drop ZH perpendicular to AZ. Hence, this line will intersect the cone’s axis, so let it intersect [that axis] at H. 
◉ Et extrahamus ex Z lineam contingentem circulum, et sit ZM, et extrahamus ex A lineam continentem cum utraque linea AZ, AH angulum acutum, et sit extra superficiem contingentem piramidem transeuntem per lineam AZ, et hoc possibile. Sit ergo AO, et extrahamus ex puncto H lineam in superficie in qua sunt AO, AH continentem cum AH angulum equalem angulo ZHA. Hec ergo linea concurret cum AO, nam duo anguli A, H sunt acuti. Concurrant ergo in O. 
◉ From Z let us extend line ZM tangent to the circle [ZU], and let us extend a line from A that forms an acute angle with both lines AZ and AH, and let it lie outside the plane tangent to the cone that passes along line AZ, which is possible. Let it be AO, then, and let us produce a line from point H within the plane containing AO and AH that forms an angle [AHO] with AH equal to angle ZHA. This line [HO] will therefore intersect AO because the two angles at A and H [i.e., OAH and AHO] are acute. So let them [i.e., AO and HO] intersect at O. 
◉ Linea ergo HO concurret cum circumferentia circuli ZU, nam angulus AHO est equalis angulo AHZ. Concurrat ergo in U, et extrahamus AU recte. Et extrahamus perpendicularem HZ ad T, et continuemus OZ, et extrahatur recte ad F, et extrahatur AZ ad E. Angulus ergo FZH erit acutus, quia linea OZ secat superficiem contingentem piramidem transeuntem per AZ. Linea ergo FZ est sub differentia communi inter superficiem OZH et superficiem contingentem, et hec differentia continet cum linea HZ angulum rectum. Angulus ergo OZH est obtusus; ergo angulus FZH est acutus. 
◉ Accordingly, line HO will intersect the circumference of circle ZU because angle AHO = angle AHZ [by construction]. So let it intersect at U, and let us extend AU in a straight line. Let us also extend perpendicular HZ to T, let us produce OZ and extend it directly to F, and extend AZ to E. Therefore, angle FZH will be acute because line OZ cuts the plane tangent to the cone and passing along AZ. Hence, line FZ lies below the common section of plane OZH and the [aforementioned] tangent plane [passing along AZ], and this common section forms a right angle with line HZ. Thus, angle OZH is obtuse, [and] so [adjacent] angle FZH is acute. 
◉ Ponatur ergo in ZF punctus F a quo extrahatur perpendicularis FE super AE, et extrahatur recte. Concurret ergo cum linea AO, nam angulus OAE est acutus. Concurrat ergo in N, et extrahatur ex E linea ED equidistans ZH linee. Erit ergo ED perpendicularis super superficiem contingentem piramidem transeuntem per AE. 
◉ So take point F on ZF, from it extend FE perpendicular to AE, and continue it in a straight line. It will therefore intersect line AO because angle OAE is acute. Let it intersect at N, then, and extend line ED from E parallel to line ZH. ED will thus be perpendicular to the plane tangent to the cone passing along AE. 
◉ Et extrahatur ex E linea equidistans linee ZM, et sit EL, et extrahatur superficies in qua sunt LE, ED. Secabit ergo superficiem piramidis et faciet sectorem, nam hec superficies est obliqua super axem AD. 
◉ Then from E draw line EL parallel to line ZM, and produce the plane containing LE and ED. It will therefore intersect the surface of the cone and will form a [conic] section, for this plane is oblique to axis AD. 
◉ Sit ergo sector BEG. Et MZ est perpendicularis super superficiem AZH, et hoc declaratum est in predictis. Ergo linea LE est perpendicularis super superficiem AED; ergo angulus AEL est rectus, et angulus AEN est rectus, et similiter angulus AED est rectus. Ergo linee LE, NE, DE sunt in eadem superficie. Ergo linea FEN est in superficie sectoris. 
◉ Let the [conic] section be BEG’. MZ is perpendicular to plane AZH [since it was constructed tangent to the circle at point Z and is therefore perpendicular to the diameter passing from Z through axis AH of the cone], and this was established earlier [in proposition 21]. Therefore, line LE is perpendicular to plane AED, so angle AEL is right, angle AEN is right, and likewise angle AED is right [all three by construction]. Consequently, lines LE, NE, and DE lie in the same plane [to which AE is perpendicular]. Hence, line FEN lies in the plane of the [conic] section [BEG’]. 
◉ Et extrahatur ex F linea equidistans linee DE, et sit FR. Hec ergo linea equidistabit linee HZ. Et extrahatur ex Z in superficie OZH linea continens cum ZT angulum equalem angulo OZT. Hec ergo linea concurret cum FR, quia secabit ZH equidistantem FR, et est in superficie eius, quia ZF est in superficie eius. Concurrat ergo in R. 
◉ From point F extend line FR parallel to line DE. Accordingly, this line will be parallel to line HZ [to which DE was constructed parallel]. In plane OZH draw a line [ZR] from Z that forms with ZT an angle equal to OZT. This line will therefore intersect FR because it will intersect ZH, which is parallel to FR [by construction] and lies in the same plane with it, since ZF lies in that plane. So let it intersect at R. 
◉ Ergo duo anguli qui sunt apud R, F sunt equales, sunt enim equales duobus angulis qui sunt apud Z. Due ergo linee RZ, FZ sunt equales. Et declaratum est quod linea FEN est in superficie sectoris, et linea FR est equidistans linee ED. Est ergo in superficie sectoris. 
◉ Accordingly, the two angles at R and F [i.e., ZRF and ZFR] are equal, for they are equal to the two angles [OZT and TZR] at Z.⁑ So the two lines RZ and FZ [within isosceles triangle ZRF] are equal. But it has been shown that line FEN lies in the plane of the [conic] section [BEG’], and line FR is parallel to line ED [by construction]. It [i.e., FR] therefore lies in the plane of the [conic] section. 
◉ Et continuemus RE. Erit ergo in superficie sectoris. Et extrahatur DE ad K, et declaratum est quod EA est perpendicularis super superficiem sectoris. Uterque ergo angulus AER, AEF est rectus, et due linee FZ, RZ sunt equales. Ergo due linee RE, FE sunt equales; ergo duo anguli ERF, EFR sunt equales. 
◉ Let us then draw RE. It will thus lie in the plane of the [conic] section. Extend DE to K, and it has been shown that EA is perpendicular to the plane of [that] section. Hence, each of the angles AER and AEF is right, and the two lines FZ and RZ are equal. Consequently, the two lines RE and FE are equal, so the two angles ERF and EFR are equal. 
◉ Ergo forma N convertetur ad R ex E, et forma O convertetur ad R ex Z. Et omnis linea extracta ex F ad aliquod punctum linee AN secabit AE. Et patet quod illa linea erit equalis linee extracte ex R, nam AE est perpendicularis super superficiem in qua sunt linee RE, FE, nam hec superficies est superficies sectoris, et due linee RE, FE sunt equales. Ergo omnes due linee extracte ex R, F ad aliquod punctum linee AE sunt equales. 
◉ Accordingly, the form of N will be reflected to R from E, and the form of O will be reflected to R from Z. Moreover, every line extended from F to some point on line AN will intersect AE. But it is clear that that line will be equal to the line extended from R [to the same point on AE where the line from F intersects it] because AE is perpendicular to the plane in which lines RE and FE lie, since this plane is the plane of the [conic] section, and the two lines RE and FE are equal. Hence, both lines extended from R and F to a given point on line AE are equal. 
◉ Patet ergo quod forma puncti quod est in AN convertetur ad R ex illo puncto quod est in AE. Et similiter de omni puncto posito in AN ultra N, si copulatum fuerit cum F, et per lineam rectam, illa linea secabit AE ultra E. Et patet quod forma puncti quod est in AN convertetur ad R ex puncto in AE. Patet ergo ex hoc quod forma linee AN, et quicquid continuatur cum ipsa, convertetur ad R a superficie piramidis ABG ex linea recta, et similiter omnis linea extracta ex A oblique super axem piramidis. 
◉ It is therefore evident that the form of [any] point on AN will be reflected to R from a point [such as] that on line AE. And the same holds for any point lying on AN beyond N; if it is connected with F by a straight line, that line will intersect AE beyond E. It is also evident that the form of a point on AN will be reflected to R from a point on AE. From this, therefore, it is evident that the form of line AN, as well as any [line] continuous with it, will be reflected to R from a straight line on the surface of cone ABG, and the same holds for every line extended from A at a slant to the cone’s axis.⁑ 
◉ Et continuemus ND. Secabit ergo circumferentiam sectoris, nam duo puncta N, D sunt in superficie sectoris, et N est extra sectorem, et D est intra sectorem. Secet igitur circumferentiam sectoris in C, et quia triangulus AOH est in eadem superficie, erit ND in superficie trianguli AOH. 
◉ Let us draw ND [in figure 6.6.22a, p. 131, abstracted from figure 6.6.22].⁑ It will therefore intersect the periphery of the [conic] section because the two points N and D lie in the plane of [that] section, and N lies outside the section[‘s periphery], whereas D lies inside the section[‘s periphery]. So let it intersect the periphery of the [conic] section at C, and since triangle AOH lies in the same plane, ND will lie in the same plane as triangle AOH. 
◉ C ergo est in superficie trianguli AOH, et duo puncta A, N sunt in superficie huius trianguli. Ergo puncta A, N, C sunt in superficie trianguli AOH. Sed puncta A, U, C sunt in superficie piramidis. Ergo puncta A, U, C sunt in differentia communi superficiei piramidis et superficiei AND. Sed hec differentia est linea recta. Ergo puncta A, U, C sunt in linea recta. 
◉ [Point] C is therefore in the plane of triangle AOH, and the two points A and N lie in the plane of this triangle. Hence, points A, N, and C lie in the plane of triangle AOH. But points A, U, and C lie on the surface of the cone. Accordingly, points A, U, and C lie on the common section of the surface of the cone and plane AND. But this common section is a straight line. So points A, U, and C lie in a straight line. 
◉ Extrahatur ergo AU recte ad C, et extrahatur RZ recte. Secabit ergo OH. Secet ergo in P. P ergo est in superficie trianguli AOH. Continuetur ergo AP, et transeat recte. Secabit ergo ND in G, et quia F est sub superficie contingente piramidem transeunte per lineam AZE, erit angulus FED acutus, et angulus DEN est obtusus. Ergo angulus ENC est acutus. 
◉ Extend AU directly to C, then, and extend RZ in a straight line. Accordingly, it will intersect OH [because O, Z, R and H all lie in the same plane of reflection]. Let it intersect at P. P therefore lies in the plane of triangle AOH. So extend AP, and let it continue in a straight line. It will therefore intersect ND at G, and since F lies below the plane tangent to the cone that passes along line [of longitude] AZE, angle FED will be acute, whereas [adjacent] angle DEN is obtuse. Hence, [interior] angle ENC [of triangle NED] is acute [because it is smaller than opposite exterior angle FED, which is acute]. 
◉ Et sit linea CZ contingens sectorem. Patet ergo, ut in figura predicta, quod angulus DCZ est obtusus et quod perpendicularis extracta ex C super CZ secat angulum DCZ, et concurret cum ED sub D. Hec ergo perpendicularis secabit ED in S. 
◉ Furthermore, let line CZ’ be tangent to the [conic] section. It is clear, then, as [shown] in an earlier proposition, that angle DCZ’ is obtuse⁑ and that the perpendicular erected on CZ’ at C cuts angle DCZ’ and will intersect ED beyond D. So this perpendicular will intersect ED at S. 
◉ Perpendicularis ergo extracta ex N super lineam contingentem sectorem secabit sectorem ultra C, scilicet, remotius ex E quam C, nam iste perpendiculares concurrent ultra circumferentiam sectoris. Perpendicularis ergo extracta ex N super lineam contingentem sectorem non secabit angulum DCZ. Erit ergo remotior ex NE quam CD, et hec perpendicularis secat ED sub D. 
◉ Hence, the perpendicular extended from N to the line tangent to the [conic] section [at the point where that perpendicular intersects the conic section] will intersect the [conic] section beyond C, that is, farther from E than C [lies from it], for these perpendiculars [i.e., NQ and ED] will intersect outside the periphery of the [conic] section [i.e., on the other side of that periphery from N]. Hence, the perpendicular extended from N to the line tangent to the [conic] section will not cut angle DCZ’. It will therefore lie farther from NE than CD [does], and this perpendicular cuts ED beyond D. 
◉ Sit ergo perpendicularis extracta ex N super lineam contingentem sectorem linea NQ. Et RE secat EN, et secat circumferentiam sectoris, et est in superficie eius, et NQ est in superficie sectoris. Si ergo RE extrahatur recte, secabit NQ. Secet ergo in Y. 
◉ So let the perpendicular dropped from N to the line tangent to the [conic] section be NQ. Also, RE intersects EN, and it intersects the periphery of the [conic] section and lies in its plane, while NQ [also] lies in the plane of the [conic] section. Hence, if RE is extended in a straight line, it will intersect NQ. Let it intersect at Y, then. 
◉ Et superficies AND secat superficiem sectoris. Quia punctum E est extra superficiem AND, nam superficies AND non est superficies sectoris, A enim est extra superficiem sectoris, quia AE est perpendicularis super superficiem sectoris, et E est in circumferentia illius, ergo ND est differentia communis superficiei AND et superficiei sectoris, et NQ concurret cum sectore ultra C. Ergo NQ est ultra superficiem AND. Y ergo est ultra lineam APG. 
◉ Plane AND intersects the plane of the [conic] section. Since point E lies outside plane AND because plane AND is not [in] the plane of the [conic] section [whereas E is], and because A lies outside the plane of the [conic] section, since AE is perpendicular to the plane of [that] section, whereas E lies on its periphery, then ND is the common section of plane AND and the plane of the [conic] section, and NQ will intersect [that] section beyond C [i.e., on the opposite side from C and E]. Hence, NQ lies outside plane AND. Y therefore lies outside line APG. 
◉ Si ergo visus fuerit in R, et linea AON fuerit in aliquo visibili, tunc P erit ymago O, et Y erit ymago N, et A videbitur in suo loco, quia est in capite piramidis. Et erit ymago linee AON linea transiens per puncta A, P, Y, sed hec linea est convexa, quia est ultra APG. 
◉ Accordingly, if the center of sight lies at R, and if line AON lies on some visible object, then P will be the image of O, Y will be the image of N, and A will appear at its [actual] location, since it lies at the vertex of the cone.⁑ And the image of line AON will be the line passing through points A, P, and Y, but this line is convex because it lies outside [straight line] APG. 
◉ Sit ergo illa linea APY, et patuit iam quod forme omnium punctorum que sunt in AN convertuntur ad R ex AE. Linee ergo radiales per quas convertuntur ille forme sunt in superficie trianguli RZE; omnes ergo ymagines linee AN sunt in hac superficie. 
◉ So let that [image]line be APY, and it has already been shown that the forms of all the points on AN are reflected to R from AE. Therefore, the radial lines according to which those forms are reflected lie in the plane of triangle RZE, so all the images of [the points on] line AN lie in that plane. 
◉ Ergo linea APY convexa est in hac superficie, et P est propinquius ad R quam Y, et erit convexitas huius ymaginis ex parte visus, et erit convexitas parva. Et dyameter huius ymaginis erit minor ipsa linea modica quantitate. Ymagines ergo linearum rectarum que extrahuntur ex capite piramidali oblique super axem comprehenduntur a visu in tali speculo convexe, et forme harum linearum convertuntur a lineis rectis ex lineis extensis in longitudine piramidis, et hoc est quod voluimus declarare. 
◉ Hence, convex line APY lies within that plane, and P lies closer to R than Y does, and the convexity of this image will be toward the center of sight, and it will [therefore] be of slight [apparent] convexity.⁑ Moreover, the [length along the] crosssection [AY] of this image will be slightly smaller than the line itself [i.e., AN of which it is the image]. Consequently, the images of straight lines that are extended from the vertex of the cone at a slant to the axis are perceived by sight as convex in such a mirror, and the forms of these lines are reflected from straight lines among the lines extended along the cone’s longitude, and this is what we wanted to prove. 
◉ Forme vero linearum equidistantium latitudini speculi piramidalis convexi convertuntur a lineis convexis in superficie speculi, et convexitas harum linearum patet ut in columpnali speculo convexo, et per illam eandem viam, et patebit similiter quod ymagines harum linearum erunt nimium convexe et manifeste sensui. Et erit centrum visus extra superficies in quibus est convexitas formarum harum linearum, et erunt dyametri ymaginum harum linearum multum minores ipsis lineis. 
◉ On the other hand, the forms of lines that are parallel to the width of a conical convex mirror are reflected from convex lines on the mirror’s surface, and the convexity of these lines is obvious, as [it is] in a convex cylindrical mirror, and for the same reason, and it will likewise be evident that the images of these lines will be quite convex and manifestly [so] to the [visual] sense. Also, the center of sight will lie outside the plane that contains the convexity of the forms of these lines, and the crosssections of the images of these lines will be considerably shorter than the lines themselves. 
◉ De lineis vero obliquis existentibus inter hos duos modos, que appropinquant in suo situ lineis extensis in longitudine piramidis habent formas parum convexas, que vero appropinquant lineis equidistantibus latitudini piramidis habent formas manifeste convexas. 
◉ As to lines that are slanted between these two extremes, however, those whose orientation approaches that of lines extended along the length of the cone have slightly convex forms, whereas those that approach lines parallel to the width of the cone have forms that are clearly convex. 
◉ Sed tamen linee tortuose que appropinquant capiti piramidis habent formas minores, et strictiores, et convexiores, que vero appropinquant basi piramidis habent formas ampliores propter illud quod declaratum fuit in speculis spericis convexis—scilicet quod quanto minus fuerit speculum tanto minores erunt circuli qui cadunt in superficie eius—et sic ymagines erunt propinquiores centro, ideo ergo erunt minores. 
◉ But also, curved lines that approach the vertex of the cone have smaller, narrower, and more convex forms, whereas those that approach the base of the cone have larger forms, according to what was demonstrated for convex spherical mirrors—i.e., that the smaller the mirror, the smaller the circles that fall on its surface—and so the images [falling on those smaller circles] will lie closer to the center [of curvature], from which it follows that they will be smaller. 
◉ Et similiter sectores qui cadunt in speculo piramidali qui sunt ex parte capitis piramidis sunt strictiores et minores, et sic ymago erit propinquior puncto in quo concurrunt perpendiculares exeuntes a linea visibili perpendiculariter super lineas contingentes sectores que sunt differentie communes, et ideo iste ymagines erunt minores. 
◉ By the same token, sections that lie on a conical mirror toward the cone’s vertex are narrower and shorter [than those that lie farther from it], and so the image [within such a section] will be nearer the point where the normals dropped from the visible line to the lines tangent to the sections, which form the common section [of the plane of reflection and the plane tangent to the mirror at that point], intersect, and so those images will be smaller. 
◉ Sectores vero qui sunt ex parte basis piramidis econverso, unde accidit quod forma comprehensa in speculo piramidali convexo erit piramidata, quod scilicet fuerit ex parte capitis speculi erit strictius, et quod ex parte basis erit amplius, et convexitas latitudinis forme erit manifesta. 
◉ On the other hand, the opposite holds for sections that lie toward the base of the mirror, so it happens that a form perceived in a convex conical mirror will take on a conical form, i.e., what lies toward the vertex of the mirror will be narrower, whereas what lies toward the base will be broader, and the convexity of a form along the width [of the mirror] will be evident. 
◉ Et accidit etiam in hiis speculis quod quanto magis res visa appropinquaverit speculo videbitur maior, et quanto magis erit remota videbitur minor. 
◉ It also happens in these [sorts of] mirrors that the closer the visible object approaches the mirror, the larger it will appear, whereas the farther away it will be, the smaller it will appear. 
◉ Fallacie ergo que accidunt in huiusmodi speculis sunt similes in omnibus dispositionibus illis que accidunt in speculis columpnalibus convexis preter quam in piramidatione forme. Et omnino forma rei vise que comprehenditur per conversionem semper assimulabitur forme superficiei speculi a qua convertitur forma, et huius causa est quod semper locus ymaginis constituitur ex forma superficiei speculi et ex loco concursus perpendicularium, ideo semper superficies speculi habet aliquam dignitatem in forma rei vise que comprehenditur in speculo. Fallacie vero composite in hoc speculo similes sunt fallaciis in speculis predictis. 
◉ Therefore, the misperceptions that occur in these sorts of mirrors are in every way like those that occur in convex cylindrical mirrors except for the conical shape of the form. And without exception the form of a visible object that is perceived by reflection will always take the shape of the surface of the mirror from which the form is reflected, and the reason for this is that the imagelocation is invariably determined by the shape of the mirror’s surface and by the place where the normals intersect, so the [shape of the] mirror’s surface always plays a role in the shape [of the image] of the visible object that is perceived in the mirror. However, the compound misperceptions [arising] in this [sort of] mirror are identical to the [compound] misperceptions [occurring] in the previously discussed mirrors [i.e., convex spherical and convex cylindrical].⁑ 
◉Capitulum septimum 
◉CHAPTER 7 
De fallaciis que accidunt in speculis spericis concavis 

◉ In hiis vero plures accidunt quam in omnibus speculis convexis et superficialibus, accidit enim in eis que in illis accidunt—scilicet debilitas lucis et coloris et diversitas situs et remotionis—nam causa huius est tantum conversio, non forma speculi. Accidit etiam in hiis speculis ex diversitate quantitatis plus quam in speculis convexis, nam in convexis, in maiori parte, res comprehendetur minor, in concavis vero quandoque comprehendetur maior, quandoque minor, quandoque secundum quod est, et hoc secundum diversitatem positionum eius ex speculo et ex visu, prout nos declarabimus in hoc capitulo. 
◉ In these [mirrors], in fact, more [misperceptions] occur than in all the convex and plane mirrors,⁑ for what occurs in the latter occurs in these as well—i.e., a weakening of light and color and a variation in orientation and distance—for it is reflection alone, not the shape of the mirror, that causes this [sort of variation]. [But] in addition, there is more variation in [image] size in these mirrors than in convex mirrors, for in convex [mirrors] an object will generally be perceived as smaller [than it actually is], whereas in concave [mirrors] it will sometimes be perceived as larger, sometimes as smaller, [and] sometimes as it actually is, and this happens according to how it changes position with respect to the mirror as well as to the center of sight, as we will demonstrate in this chapter. 
◉ Accidit etiam in hiis speculis quod unum visibile videatur duo, et tria, et quattuor, et non est ita in speculis superficialibus et convexis, unum enim visibile non comprehenditur in illis nisi unum, in concavis vero non. 
◉ It also happens in these mirrors that a single visible object may appear as two, or three, or four, and this is not the case in plane and convex mirrors, for in those [kinds] a single visible object is perceived only singly, whereas in concave [mirrors, such is] not [the case]. 
◉ Item ordinatio partium rei vise comprehenditur in speculis convexis et superficialibus secundum quod est; in spericis vero concavis in pluribus sitibus alio modo, et hec duo: scilicet comprehensio unius ut unius et comprehensio ordinationis partium secundum quod est, non habent aliquam deceptionem in speculis convexis spericis, et cum in hiis accidit deceptio in speculis spericis concavis, patet quod nichil comprehenditur in huiusmodi speculis nisi cum fallacia, aut semper, aut in aliqua hora secundum diversitatem positionis. 
◉ Furthermore, the arrangement of the visible object’s parts is perceived in convex and plane mirrors as it actually is, whereas in spherical concave [mirrors it is perceived] otherwise in several situations,⁑ and this in two ways: namely, in convex spherical mirrors there is no deception in the perception that a single thing is single and the perception of the arrangement of its parts according to how it actually is, and since there is deception in regard to these aspects in spherical concave mirrors, it is clear that nothing is perceived in these mirrors without deception, either invariably or at some time according to variation in the position [of the object visàvis the mirror as well as the center of sight]. 
◉ Debilitas vero lucis et coloris et diversitas positionis et distantia accidunt in hiis speculis sicut in aliis semper, et in omni positione. Quantitas vero, et forma, et numerus habent deceptiones in hiis speculis in aliquibus sitibus, prout declarabimus. 
◉ However, weakening of light and color as well as change in position and distance occur in these mirrors just as [they] invariably [occur] in the others, and they do so in every situation. But size, shape, and number are subject to deception in these mirrors in some situations, as we will demonstrate. 
◉ De numero vero declaratum est in capitulo de ymagine quod unum visum in speculis spericis concavis habet unam ymaginem, et duas, et tres, et quattuor, et quod forma rei vise semper comprehenditur in loco ymaginis. Verum unum visum comprehensum in speculis spericis et concavis forte comprehendetur unum, et forte duo, et forte tria, et forte quattuor, quod non accidit in speculis convexis et superficialibus. 
◉ Concerning number it has been shown in chapter [2, book 5] on image [formation] that in concave spherical mirrors one object has one, two, three, or four images, and that the form of a visible object is always perceived at its [appropriate] imagelocation. However, one object perceived in concave spherical mirrors may be perceived as one, perhaps as two, perhaps as three, and perhaps as four, which does not happen in convex and plane mirrors. 
◉ De ordinatione vero partium rei vise dictum est etiam in capitulo de ymagine quod forma unius puncti convertitur ex circumferentia circuli, et quod visibilia quorum ymagines sunt retro vel post visum, et ante, et in centro visus apparent dubia, non certificata, et quod est huiusmodi non habet ordinationem partium sicut ipsa res visa habet. Et hoc etiam est in hiis speculis aliter quam sit in speculis convexis et superficialibus. Cause autem huius rei declarate sunt in capitulo de ymagine. 
◉ As to the arrangement of the visible object’s parts, it has also been claimed in chapter [2, book 5] on image [formation] that the form of a single [object]point is reflected from the circumference of a [great] circle [on the mirror’s surface] and that visible objects whose images lie beyond or behind the center of sight, in front of it, or at the center of sight [itself] appear blurred and not clear, and anything of this sort does not have the arrangement of parts that the visible object itself has. And here, as well, what obtains in these mirrors is other than what obtains in convex and plane mirrors. But the reasons for this phenomenon have been discussed in the chapter on image [formation].⁑ 
◉ Restat ergo declarare quod illud quod comprehenditur in hiis speculis forte comprehendetur maius, et forte minus, et forte equale, et quod in quibusdam positionibus comprehendetur conversum, et in quibusdam erectum, et quod rectum in huiusmodi speculis comprehenditur concavum et convexum et rectum, et quod convexum et concavum comprehenduntur etiam aliter quam sint. Et hec etiam sunt ex diversitate ordinationis partium rei vise, et nos declarabimus hoc hoc modo. 
◉ It thus remains [for us] to show that what is perceived in these mirrors may be perceived larger, smaller, or the same size [as the object itself], and that in certain situations it may be perceived inverted and in others erect, and that a straight object is perceived as concave, convex, or straight in mirrors of this sort, and that convex and concave objects are also perceived other than they [actually] are [in this sort of mirror]. And these [misperceptions] also arise from a variation in the arrangement of the visible object’s parts, and we will demonstrate this in the following way. 
◉ [PROPOSITIO 23] Sit ergo speculum spericum concavum in centro A [FIGURE 6.7.23, p. 317], et secetur superficie equali transeunte per centrum, et faciat circulum BG, et extrahatur in ipsa linea, quocumque modo sit, et dividatur in duo equalia in O. 
◉ [PROPOSITION 23] Accordingly, let there be a concave spherical mirror centered on A [in figure 6.7.23, p. 132], let it be bisected by a plane passing through its center, let it form [great] circle BG, let a line [AU] be extended within it at random, and let it be bisected at O. 
◉ Et ponatur A centrum, et in distantia AO faciamus circulum, et sit EZ. Et ponatur in linea OU punctum T casualiter, quocumque modo sit, et ex T extrahantur linee TN, TM recte super lineam AU. Et extrahantur ex T linee TE, TZ tangentes circulum EZ, et continuemus AE, AZ, et transeant ad B, G. Et continuemus TB, TG, et extrahamus BM equidistantem ad AT, et GN etiam equidistantem AT, et continuemus AM, AN et extrahantur recte. 

◉ Quia ergo AO est sicut OU, erit AE sicut EB, et AZ sicut ZG, et quia TE tangit circulum EZ, erit TE perpendicularis super BA, et similiter TZ perpendicularis super AG. Linea ergo BT est sicut TA, et TG sicut TA, et angulus TBA sicut angulus TAB, et angulus TGA sicut angulus TAG. Et quia BM est equidistans AT, erit angulus MBA sicut angulus BAT. Ergo angulus MBA est sicut angulus ABT, et similiter angulus TGA est sicut angulus AGN. 
◉ Therefore since AO = OU [by construction], AE = EB, and AZ = ZG, and because TE is tangent to circle EZ, TE will be perpendicular to BA, and likewise TZ [will be] perpendicular to AG. Hence, line BT = [line] TA [by Euclid, I.4], [line] TG = [line] TA, angle TBA = angle TAB [within isosceles triangle TBA], and angle TGA = angle TAG [within isosceles triangle TGA]. And since BM is parallel to AT, angle MBA = [alternate] angle BAT. Therefore, angle MBA = angle ABT, and likewise angle TGA = angle AGN. 
◉ Cum ergo visus fuerit in T, et M, N fuerit in aliquo visibili, tunc forma M extendetur per lineam MB et convertetur per BT, et forma N extendetur per NG et convertetur per GT. Visus ergo T comprehendet puncta M, N ex punctis B, G, et lineam MN ex arcu BG. 
◉ So when the center of sight is at T, and when M and N lie on some visible object, the form of M will be extended along line MB and will be reflected along BT, and the form of N will be extended along NG and will be reflected along GT. The center of sight at T will therefore perceive points M and N [at locations] beyond points [of reflection] B and G, and [so it will perceive the entire image of] line MN beyond arc BG.⁑ 
◉ Et quia TE est perpendicularis super AB, erit angulus ABT acutus. Sed angulus MBA est sicut angulus ABT. Ergo TB est maior BM; ergo AT est maior BM, et sunt equidistantes. Ergo TB concurret cum AM. Concurrant ergo in F. F ergo est ymago M, et sic declarabitur quod TG concurret cum AN. Concurrat ergo in Q. Q ergo erit ymago N. 
◉ Also, since TE is perpendicular to AB, angle ABT will be acute. But angle MBA = angle ABT. Thus, TB > BM, so AT > BM, and they [i.e., lines AT and BM] are parallel. Consequently, [line of reflection] TB will intersect [cathetus] AM. Let them intersect at F, then. F is thus the image of M, and it will be demonstrated equivalently that [line of reflection] TG will intersect [cathetus] AN. Let it intersect at Q, then. Q will thus be the image of N. 
◉ Et continuemus FQ, que est dyameter ymaginis MN, et quia TE, TZ sunt equales, erunt anguli TAB, TAZ equales, et erunt linee TB, TG equales, et linee BM, GN equales, et linee AM, AN equales. Et proportio AF ad FM sicut proportio AT ad BM, et proportio AF ad FM est sicut proportio AT ad GN, et proportio AT ad BM; ergo proportio AF ad FM est sicut proportio AQ ad QN, et AM est sicut AN. Ergo AF est sicut AQ; ergo FQ equidistat MN. Ergo FQ est maior MN. Sed FQ est dyameter ymaginis MN. Ergo si visus fuerit in T et MN fuerit in aliquo visibili, tunc visus comprehendet formam maiorem quam sit. 
◉ Let us then connect FQ, which is the crosssection of the image of MN, and since TE and TZ are equal, angles TA[E]B and TAZ[G] will be equal, lines TB and TG will be equal, lines BM and GN [will be] equal, and lines AM and AN [will be] equal. Moreover [given the similarity of triangles AFT and MFB], AF:FM = AT:BM, and AF:FM = AT:GN = AT:BM [because GN = BM], so AF:FM [= AT:GN] = AQ:QN, and AM = AN. Hence, AF = AQ, so FQ is parallel to MN. Thus, FQ > MN. But FQ is the crosssection of the image of MN. Accordingly, if the center of sight is at T and MN lies on some visible object, the eye will perceive its form as larger than [objectline MN] is. 
◉ [PROPOSITIO 24] Item iteremus circulum BG [FIGURE 6.7.24, p. 317], et lineam AT, et lineas AB, AG, TB. Et super punctum T sit perpendicularis super superficiem circuli BG, et sit TK, et continuemus KA, KB, KG. Superficies ergo KBA, KGA sunt secantes speram super suum centrum perpendiculariter super superficies tangentes ipsam. Ex ipsis ergo convertitur forma, et due differentie communes inter has duas superficies et speram sunt circuli magni a quorum circumferentia convertuntur forme. 
◉ [PROPOSITION 24] Now [in figure 6.7.24, p. 132] let us duplicate circle BG, line AT[U], and lines AB, AG, and TB [as given in figure 6.7.23]. Let TK be perpendicular to the plane of circle BG at point T, and let us draw KA, KB, and KG. Thus, planes KBA and KGA intersect the sphere [of the mirror] at its center perpendicular to [the appropriate] planes tangent to its surface.⁑ Within these [planes], then, the form [of any given visible object] is reflected, and the two common sections between these two planes and [the surface of] the sphere form great circles from whose circumference the forms are reflected. 
◉ Et extrahamus BM in superficie BKA equidistantem AK, et sit minor quam AK. Et continuemus AM, et extrahatur recte, et extrahatur KB donec concurrant in F. Et extrahatur NG in superficie KGA, et sit equidistans AK, et ponatur equalis BM. Et continuemus AN, extrahatur recte, et extrahatur KG recte donec concurrant in Q. Et continuemus MN, FQ. 
◉ Let us then draw BM parallel to AK in plane BKA, and let it be shorter than AK. Let us draw AM, and let it be extended in a straight line, and extend KB until they intersect at F. Then draw NG in plane KGA, let it be parallel to AK, and assume it is equal to BM. Let us connect AN, let it be extended in a straight line, and extend KG in a straight line until they intersect at Q. Then let us connect MN and FQ. 
◉ Quia ergo BT est sicut TA, erit BK sicut KA, et GK sicut KA. Ergo BK est sicut GK, et angulus KBA est sicut angulus KGA, et angulus KAB est sicut angulus KBA. Et similiter angulus KGA est sicut angulus KAG; ergo angulus ABM est sicut angulus ABK, et angulus AGN est sicut angulus AGK, et erit angulus ABM sicut angulus AGN. Et linea BM erit sicut linea GN. Tunc linea AM erit sicut linea AN; tunc AF erit sicut linea AQ. Tunc due linee FQ, MN erunt equidistantes; tunc FQ erit maior linea MN. 
◉ Accordingly, since BT = TA [as concluded in the previous proposition], BK = KA [by Euclid, I.4], and GK = KA [by Euclid, I.4]. Hence, BK = GK, angle KBA = angle KGA, and angle KAB = angle KBA. Likewise, angle KGA = angle KAG, so angle ABM [which is alternate to angle KAB] = angle ABK [which = angle KAB], angle AGN [which is alternate to angle KAG] = angle AGK [which = angle KAG], and angle ABM = angle AGN [since both are equal to equal angles KBA and KGA]. In addition, line BM = line GN [by construction]. Thus, line AM = line AN; so [line] AF = line AQ. The two lines FQ and MN will therefore be parallel, so [line] FQ > line MN. 
◉ Tunc quando fuerit visus super punctum K, et fuerit linea MN in aliquo visibili, tunc forma M extendetur super lineam MB et convertetur per lineam BK in superficie circuli transeuntis per puncta B, A, K, et forma puncti N extendetur super lineam NG et convertetur super lineam GK in superficie circuli transeuntis per puncta G, A, K. 
◉ Hence, when the center of sight lies at point K, and when line MN lies on some visible object, the form of M will be extended along line MB and will be reflected along line BK in the plane of the circle passing through points B, A, and K, whereas the form of point N will be extended along line NG and will be reflected along line GK within the plane of the circle passing through points G, A, and K. 
◉ Et erit punctum F ymago puncti M, et punctum Q erit ymago puncti N, et erit linea FQ dyameter ymaginis MN. Et iam declaravimus quod linea FQ est maior linea MN; tunc quando fuerit visus super punctum K, et fuerit linea MN in aliquo visibili, tunc visus apprehendet formam linee MN super lineam FQ. Tunc comprehendet formam maiorem re visa. 
◉ And [so] point F will be the image of point M, while point Q will be the image of point N, and line FQ will be the crosssection of the image of [the entire line] MN. But we have already demonstrated [in paragraph 7.16] that line FQ > line MN, so when the center of sight is at point K, and when line MN lies on some visible object, the eye will apprehend the form of line MN on line FQ. Therefore, it will perceive the form [of the visible object] as larger than the visible object [itself]. 
◉ Et sic, si revolverimus item totam figuram in circuitu linee AU, ipsa immobili, tunc punctus K faciet circulum perpendicularem super lineam AU, et sic omne punctum ultra illum punctum illius circuli habebit situm respectu linee comparis linee MN sicut est situs K respectu MN. 
◉ Accordingly, if we rotate the entire figure around line AU, while keeping [AU] itself stationary [to form the axis of rotation], point K will produce a circle that is perpendicular to line AU, and so every point beyond that point on that circle will be situated with respect to a line equivalent to line MN as K is situated with respect to MN. 
◉ Si ergo visus fuerit in aliquo puncto circumferentie huius circuli et linea compar linee MN fuerit in superficie alicuius rei vise, tunc visus comprehendet formam illius linee maiorem. Et similiter si extraxerimus TK recte et posuerimus in ipsa aliquod punctum preter K, et extraxerimus semper ab illo puncto, quod est quasi punctum K, erit modus eius sicut modus puncti K. 
◉ Consequently, if the center of sight lies at any point on the circumference of this circle, and if a line equivalent to line MN lies on the surface of some visible object [that is similarly disposed], the eye will perceive the form of that line [as] larger [than the line itself]. Likewise, moreover, if we extend TK in a straight line and take some point on it other than K [as a center of sight], and if we extrapolate at every stage from that point, which is equivalent to point K, the case will be like the case for point K. 
◉ Ex hiis ergo duabus figuris patet quod in spericis speculis concavis multa et ex multis sitibus comprehenduntur maiora. 
◉ On the basis of these two propositions [i.e., 23 and 24], therefore, it is evident that in concave spherical mirrors many objects are perceived [as] larger [than they actually are] in many situations. 
◉ [PROPOSITIO 25] Item sit speculum spericum concavum AB circa centrum E [FIGURE 6.7.25, p. 318], et extrahamus superficiem transeuntem per E, et faciat circulum AB. Et extrahamus ex E lineam EZ, quocumque modo fuerit, usque ad G, et ex G extrahamus GD perpendicularem super superficiem circuli AB, et in ipsa signemus punctum D, quocumque modo fuerit. Et continuemus DE, et extrahamus ipsam usque ad O, et extrahamus EB ita quod contineat cum ED angulum obtusum, et extrahamus EA ita quod contineat cum ED angulum equalem angulo DEB. Et continuemus DA, DB. Sic ergo superficies duorum triangulorum DAE, DBE secant se super lineam DE, et duo anguli acuti DBE, DAE erunt equales. 
◉ [PROPOSITION 25] To continue, let AB [in figure 6.7.25, p.133] be a concave spherical mirror centered on E, and let us produce a plane passing through E, and let it form [great] circle AB [on the sphere]. Let us extend line EZ randomly from E to G, and from G let us drop GD perpendicular to the plane of circle AB, and let us mark point D on it at random. Then let us connect DE and extend it to O, let us produce EB so that it forms an obtuse angle [DEB] with ED, and let us produce EA so that it forms an angle [AED] with ED equal to angle DEB. Let us then connect DA and DB. Accordingly, the planes of the two triangles DAE and DBE intersect one another along line DE, and the two acute angles DBE and DAE will be equal. 
◉ Et extrahamus ex B lineam in superficie trianguli DBE continentem cum EB angulum equalem angulo DBE. Hec ergo linea concurrit cum linea DE, quia angulus BEO est acutus, et angulus qui est apud B est acutus. Concurrat ergo in O. 
◉ Now from B in the plane of triangle DBE let us produce a line [BO] forming an angle [EBO] with EB equal to angle DBE. Hence, this line intersects line DE, since angle BEO is acute, and the angle [EBO] at B is acute. So let it intersect at O. 
◉ Et extrahamus etiam ex A lineam in superficie trianguli DAE continentem cum AE angulum equalem angulo DAE. Concurrat ergo cum DE in O, quia duo anguli AEO, BEO sunt equales, et anguli qui sunt apud duo puncta A, B sunt equales. 
◉ From A let us also produce a line [AO] in the plane of triangle DAE that forms an angle [EAO] with AE equal to angle DAE. So let it intersect DE at O because the two angles AEO and BEO are equal [by construction], and because the angles at the two points A and B [i.e., EAO and EBO] are equal [by construction]. 
◉ Et extrahamus ET ita quod contineat cum EB angulum rectum, et extrahamus TE in parte E, et BO in parte O, et concurrant in H, et erit TE equalis EH. Et similiter extrahamus EK ita quod contineat cum EA angulum rectum, et extrahamus illam in parte E, et extrahamus AO, et concurrant in L. Sic ergo KE erit equalis EL. 
◉ Let us then produce ET so that it forms a right angle with EB, and let us extend TE in the direction of E and BO in the direction of O, and let them intersect at H, and [so] TE = EH [insofar as triangles TEB and HEB are equal, by Euclid, I.26]. Let us likewise produce EK so that it forms a right angle with EA, let us extend it in the direction of E, and let us extend AO, and let them intersect at L. Therefore, KE = EL. 
◉ Et continuemus TK, LH. Erunt ergo equales. Si ergo visus fuerit in D et LH fuerit in aliquo visibili, tunc D comprehendet LH in speculo AB, et erit T ymago H, et K ymago L, et sic TK erit dyameter ymaginis LH, et est ei equalis. 
◉ Let us then connect TK and LH. They will thus be equal [because they lie within triangles KTE and HLE that are equal, insofar as KE = EL, TE = EH, and angle KET = vertical angle HEB]. Hence, if the center of sight lies at D, and if LH lies on some visible object, then D will perceive LH in mirror AB, and T will be the image of H [whose form is reflected from point B], K the image of L [whose form is reflected from point A], and so TK will be the crosssection of the image of LH, and it is equal to it. 
◉ Si ergo revolverimus totam figuram, HL immobili, tunc D faciet circulum, et si visus fuerit in aliquo puncto circumferentie illius, poterit comprehendere aliquod visibile compar linee LH, et erit ymago eius equalis ei. Et similiter si visus fuerit in O, et res visa fuerit TK, erit ymago equalis rei vise. 
◉ Consequently, if we rotate the entire figure, leaving HL stationary [as the axis of rotation], D will produce a circle, and if the center of sight lies at any given point on its circumference, it can perceive some visible object equivalent to line LH, and the image will be equal [in size] to it. And likewise, if the center of sight lies at O and TK is the visible object, the image will be the same size as the visible object. 
◉ Sed tamen cum res visa fuerit LH, et visus fuerit D, et fuerit ymago TK, ymago erit conversa; si H fuerit in dextro, erit T in sinistro, et si H fuerit in sinistro, T erit in dextro, et si H fuerit supra lineam, erit T infra lineam, et similiter L. 
◉ But yet, if the visible object is LH, if the eye is at D, and if TK is the image, the image will be inverted; [for] if H lies on the right [of object HL from D’s point of view its image] T will lie on the left [of image TK from that same point of view], whereas if H lies to the left, T will lie to the right, and if H lies above the line, T will lie below the line, and the same for L. 
◉ Et si res visa fuerit TK, et visus fuerit O, et ymago fuerit LH, forma erit recta, nam ymago LH erit retro post visum, et comprehendetur ante rem visam, sicut declaravimus in capitulo ymaginis quinti tractatus, et visus comprehendet H, quod est ymago T, in linea BO, et L quod est ymago K, in LO. 
◉ Moreover, if the visible object is TK, if the center of sight is at O, and if the image is LH, the form will be erect, for image LH will lie beyond the center of sight, and it will be perceived ahead of the visible object, as was shown in chapter [2] on image [formation] in the fifth book, and the eye will perceive H, which is the image of T, along line BO, and L, which is the image of K, along LO.⁑ 
◉ Patet ergo quod in speculis concavis comprehenditur res visa quandoque equalis sibi. 
◉ It is therefore clear that an object is sometimes perceived in concave [spherical] mirrors the same size as it [actually] is. 
◉ [PROPOSITIO 26] Item extrahamus BH recte, et in ipsa signemus R, et continuemus RE. Sic ergo angulus REB erit obtusus. 
◉ [PROPOSITION 26] Now let us continue BH [in figure 6.7.26, p. 133] in a straight line and mark R on it[s extension], and let us connect RE. Angle REB will therefore be obtuse [since it is larger than HEB, which is a right angle, by construction]. 
◉ Et extrahamus RE ad N. Sic ergo RB erit maior quam BN. Et proportio RB ad BN est sicut proportio RE ad EN; ergo linea RE est maior quam EN. 
◉ Let us then extend RE to N. Hence, RB > BN [because BT = HB in isosceles triangle HBT, and BN < HT, while RB > HB]. Moreover, RB:BN = RE:EN [by Euclid, VI.3, since EB bisects angle NBR], so line RE > [line] EN. 
◉ Et extrahamus AL recte, et sit AM equalis BR. Et continuemus ME, et transeat usque ad U. Erit ergo ME maior quam EU. Et continuemus MR, NU. Erit ergo MR maior quam NU. 
◉ Let us also extend AL in a straight line [to M], and let AM = BR. Let us connect ME, and let it continue to U. Thus, ME > EU. Then let us connect MR and NU. MR will therefore be longer than NU. 
◉ Si ergo MR fuerit in aliquo visibili, et visus fuerit in D, erit NU dyameter ymaginis MR, et NU est minor quam MR. Et si visus fuerit in O, et NU fuerit in aliquo visibili, erit MR ymago NU, et est maior quam NU. 
◉ Accordingly, if MR lies on some visible object, and if the center of sight is at D, NU will be the crosssection of the image of MR, and NU < MR. On the other hand, if the center of sight is at O, and if NU lies on some visible object, MR will be the image of NU, and it is longer than NU. 
◉ Sed cum MR fuerit visibile, et fuerit ymago NU, tunc ymago erit conversa, et si NU fuerit visibile et MR fuerit ymago, ymago erit recta, nam ymago, si fuerit ultra visum, videbitur ante, et omne punctum ymaginis videbitur in linea in qua est de lineis radialibus. 
◉ But if MR is the visible object, and if NU is its image [as viewed from D], then the image will be inverted, whereas if NU is the visible object and MR is its image [as viewed from O], the image will be correctly oriented, for if it lies beyond the center of sight, that image will appear ahead [of the object], and every point on the image will appear along a specific line among the [corresponding] radial lines.⁑ 
◉ [PROPOSITIO 27] Item signemus in linea OH punctum Q. Et continuemus QE, et transeat ad C. Et sit OF equalis OQ, et continuemus EF, et transeat ad I. Erunt ergo due linee CE, EI maiores duabus lineis EF, QE, et erit linea CI maior quam linea FQ. 
◉ [PROPOSITION 27] To continue, let us mark point Q on line OH [in figure 6.7.27, p. 134]. Let us connect QE, and let it continue to C. Let OF = OQ, and let us connect EF and let it continue to I. The two lines CE and EI will thus be longer than the two lines EF and QE, and [so] line CI > line FQ [in similar triangles EIC and QEF]. 
◉ Si ergo visus fuerit in O, et CI in aliquo visibili, erit FQ ymago CI, et FQ est minor quam CI. Et FQ videbitur super duas lineas AO, OB. Erit ergo forma ante visum et minor quam res visa, et erit recta. 
◉ Hence, if the center of sight is at O, and if CI lies on some visible object, FQ will be the image of CI, and FQ < CI. Moreover, FQ will appear along the two lines AO and OB. Therefore, the form [of CI] will lie in front of the center of sight and will be smaller than the visible object [itself], and it will be properly oriented. 
◉ Et si visus fuerit in D, et FQ fuerit in aliquo visibili, erit CI ymago FQ. Et est maior quam FQ, et erit forma ante visum conversa. 
◉ But if the center of sight is at D, and if FQ lies on some visible object, CI will be the image of FQ. It is longer than FQ, and [so its] form will be inverted in front of the center of sight. 
◉ Patet ergo quod in speculis concavis comprehenditur forma rei vise minor, et maior, et equalis. 
◉ So it is evident that in concave [spherical] mirrors the form of a visible object is perceived as smaller [than the object itself], larger [than that object], or the same size [as the object]. 
◉ [PROPOSITIO 28] Item sit speculum concavum AB [FIGURE 6.7.28, p. 319], et centrum G, et habeat superficiem equalem transeuntem per centrum, et faciat circulum AB. Et extrahamus lineam GD, quomodocumque sit, et transeat in parte G ad E, et sit visus in E, et sit T in superficie visus. 
◉ [PROPOSITION 28] Now let AB [in figure 6.7.28, p. 135] lie [on] a concave [spherical] mirror [with] G its center, let that mirror be bisected by a plane passing through its center, and let it form [segment] AB [of a great] circle. Let us extend line GD at random, let it pass to E on the side of G, let the center of sight be at E, and let T lie on the surface of the eye. 
◉ Et extrahamus TH perpendiculariter super lineam ED, et sit ZT equalis TH, et comprehendat E punctum H ex A. Sic ergo erunt duo puncta A, H a duobus lateribus puncti G, nam si in eodem essent, tunc linea que exierit a speculo ad A non divideret angulum quem continent due linee radiales. 
◉ Then let us draw TH perpendicular to line ED, let ZT = TH, and let [the center of sight at] E perceive [the form of] H [by reflection] from A.⁑ Consequently, the two points A and H will lie on opposite sides of point G, for if they lay on the same side, the line extending from the mirror to A would not cut the angle that the two radial lines [of incidence and reflection] form.⁑ 
◉ Et extrahamus lineas EA, AH, GA, GH, et transeat GH recte ad K. Duo ergo anguli qui sunt apud A erunt equales, et erit K ymago H. 
◉ Let us then draw lines EA, AH, GA, and GH, and let GH extend in a straight line to K. Hence, the two angles at A [i.e., HAG and GAE] will be equal [by construction], and K [where cathetus HGK and line of reflection EA intersect] will be the image of H. 
◉ Et sit arcus BD equalis arcui DA, et continuemus lineas EB, BZ, BG, et extrahamus ZG ad L. Erunt ergo duo anguli apud B equales, et comprehendetur Z a visu ex B, et erit L ymago Z. 
◉ Let arc BD = arc DA, let us draw lines EB, BZ, and BG, and let us extend ZG to L. Thus, the two angles at B [i.e., ZBG and GBE] will be equal, and [the form of] Z will be perceived by the center of sight [according to reflection] from B, and L will be the image of Z. 
◉ Et continuemus KL. Erit ergo KL dyameter ymaginis ZH, et quia ZTH est perpendicularis super DE, et ZT est equalis TH, erunt due linee EA, AH equales duabus lineis EB, BZ, et duo anguli apud A sunt equales duobus angulis apud B, et linea GH est equalis linee ZG. 
◉ Let us connect KL. KL will therefore be the crosssection of the image of ZH, and since [objectline] ZTH is perpendicular to DE [by construction], and since ZT = TH [also by construction], the two lines EA and AH will be equal [respectively] to the two lines EB and BZ, the two angles [HAG and GAE] at A are equal to the two angles [ZBG and GBE] at B, and line GH = line ZG. 
◉ Ergo due linee AG, GH sunt equales duabus lineis BG, GZ, et basis AH est equalis basi BZ. Ergo angulus AHG est equalis angulo BZL, et angulus HAK est equalis ZBL. Ergo HK est equalis ZL, et linea HG est equalis ZG; ergo GK est equalis GL. Ergo KL est equidistans ZH. 
◉ Therefore, the two lines AG and GH are equal [respectively] to the two lines BG and GZ, and base AH [in triangle AGH] = base BZ [in triangle BGZ]. Consequently, angle AHG = angle BZL, and angle HAK = angle ZBL. Hence, HK = ZL, and line HG = [line] ZG, so [remainder] GK [of line HK] = [remainder] GL [of line ZL, from which it follows that the two triangles HGZ and GLK are similar and isosceles]. KL is thus parallel to ZH. 
◉ Item angulus HGA est obtusus, et duo anguli apud A sunt equales; ergo linea GH est maior linea GK, et similiter ZG est maior quam GL. Linea ergo KL est minor quam ZH. Sed KL est dyameter ymaginis ZH. Linea ergo ZH videbitur minor quam sit secundum veritatem. Et linea ZH est superficies faciei aspicientis. 
◉ Moreover, angle HGA is obtuse, and the two angles [HAG and GAE] at A are equal [by construction], so line GH > line GK, and likewise ZG > GL.⁑ Hence, line KL < [line] ZH [because of the similarity of isosceles triangles HGZ and GLK]. But KL is the crosssection of the image of ZH. Therefore, line ZH will appear shorter than it actually is. Moreover, line ZH is [on] the viewer’s face [insofar as it is a crosssection of the eye that faces the mirror]. 
◉ Si ergo revolverimus circulum ad B, EG immobili, in circuitu ED, fiet circulus, et fiet ex duobus punctis A, B circulus in superficie speculi. Et erit situs visus E respectu cuiuslibet comparis linee ZH ex illo circulo quam signat ZH, et ex omni arcu compari arcui AB ex portione quam dividit circulus quem signant duo puncta A, B sicut est situs quem visus E habet ex linea ZH et ex arcu AB. Et similiter declarabitur si posuerimus lineam maiorem quam ZH aut minorem. 
◉ Therefore, if we rotate the circle at B [i.e., arc BDA] around ED, leaving EG[D] stationary [as the axis of rotation], it will produce a circle, and it will produce a circle on the mirror’s surface from the two points A and B. In addition, the position of center of sight E with respect to any line equivalent to ZH on that circle marked off by ZH and with respect to any arc equivalent to arc AB on the segment [of the circle] that the two points A and B mark off on that circle will be equivalent to the position that center of sight E has with respect to line ZH and arc AB. And the proof will be the same whether we suppose the [object]line to be longer or shorter than ZH. 
◉ Patet ergo ex hiis omnibus quod dyameter superficiei faciei aspicientis comprehenditur in speculo concavo minor quam sit. Sequitur ergo quod si visus fuerit in E, tunc aspiciens comprehendet suam formam in tali speculo minorem quam sit, et quia K est ymago H, et L est ymago Z, erit ymago conversa. 
◉ From all of these conclusions, it is clear that the crosssection of the surface of the viewer’s face is perceived [to be] smaller than it [actually] is in the concave [spherical] mirror. So it follows that, if the center of sight lies at E, the viewer will perceive his face in such a mirror as smaller than it is, and since K is the image of H, while L is the image of Z, the image will be inverted. 
◉ Et sic visus E comprehendet suam formam, scilicet quod est in dextro comprehendet etiam in sinistro et sursum deorsum, et econverso. Et similiter si visus fuerit in quolibet puncto inter quod et superficiem speculi fuerit centrum speculi, comprehendet suam formam conversam, et hoc est quod voluimus. 
◉ Accordingly, the center of sight at E will perceive the viewer’s form as such, i.e., it will perceive what lies to the right to the left, and [what lies] below above, and viceversa. By the same token, if the center of sight lies at any point such that the center of [the mirror’s] curvature lies between it and the mirror’s surface, it will perceive its [viewer’s] form inverted, and this is what we wanted [to demonstrate].⁑ 
◉ Patet ergo ex hiis quattuor figuris quod in speculo concavo quandoque comprehenditur maior, quandoque minor, quandoque equalis, et nunc recta, et nunc conversa. 
◉ It is therefore evident from these four propositions [i.e., 2528] that in a concave [spherical] mirror [an object] is sometimes perceived as larger, sometimes smaller, and sometimes the same size [as the object itself], and [it] sometimes [appears] properly oriented, sometimes inverted. 
◉ Et in capitulo de ymagine diximus quod in speculo concavo quandoque ymago erit una, quandoque due, et quandoque tres, et quandoque quattuor, et hoc idem accidit in hiis predictis. 
◉ Moreover, in chapter [2, book 5] on image [formation], we explained that in a concave [spherical] mirror the image will sometimes be single, sometimes double, sometimes triple, and sometime quadruple, and this same phenomenon occurs in the situations just discussed. 
◉ Illud ergo quod habet ymaginem se maiorem forte habebit alias minores et equales, et quod habet minorem ymaginem forte habebit alias maiores et equales, et quod habet equalem forte maiorem et minorem, et quod rectum videtur forte videbitur sub alia ymagine conversum, et econverso. Restat ergo declarare formas eorum que comprehenduntur in huiusmodi speculis. 
◉ Hence, whatever yields an image that is larger than itself may yield others that are smaller or the same size, whereas whatever yields a smaller image may yield others larger or the same size, and whatever yields an image the same size [as itself may yield] a larger or smaller [one], and whatever appears upright [according to one image] may appear inverted according to another image, and viceversa. So it remains to analyze the forms of those things that are perceived in these sorts of mirrors. 
◉ [PROPOSITIO 29] Sit ergo speculum spericum AB [FIGURE 6.7.29, p. 320], et extrahamus in ipso speculo superficiem equalem transeuntem per centrum, et faciat circulum AB circa centrum E. Et extrahamus in hoc circulo duos dyametros se secantes AEO, BED, et speculum non excedat arcum BADO. Et ponamus in BE punctum Z, quomodocumque sit, et ponamus in linea AE punctum K, et sit AK maior quam KE. Et continuemus ZK, et transeat ad F. Et continuemus EF, et sit angulus EFG equalis angulo EFZ. 
◉ [PROPOSITION 29] Accordingly, let AB [in figure 6.7.29, p. 136] be a [concave] spherical mirror, let us produce a plane bisecting that mirror through the center, and let it form [great] circle AB centered on E. In this circle let us draw two intersecting diameters, AEO and BED, and let the mirror not extend past arc BADO. Let us then select point Z at random on BE, let us select point K on line AE, and let AK > KE. Then let us connect ZK, and let it continue to F. Let us also draw EF, and let angle EFG = angle EFZ. 
◉ Quia ergo FK est maior quam KA, et KA est maior quam KE, erit FK maior quam KE. Angulus ergo FEK est maior angulo EFK; est ergo maior angulo EFG. Linea ergo FG concurret cum linea KE. Concurrant ergo in G. Due ergo linee ZF, FG convertuntur per angulos equales; K ergo est ymago G, si visus fuerit in Z. 
◉ Thus, since FK > KA, and since KA > KE [by construction], FK > KE.⁑ Angle FEK is therefore greater than angle EFK [by Euclid, I.19], so it is greater than angle EFG. Hence, line FG will intersect line KE.⁑ Let them intersect at G, then. Consequently, the two lines ZF and FG are reflected at equal angles [ZFE and GFE], so K [where cathetus GEK and line of reflection ZKF intersect] is the image of G if the center of sight is at Z. 
◉ Et extrahamus lineam ZLH, quomodocumque sit, et continuemus EH, HG, ZG, et extrahamus FE usque ad M. Proportio ergo ZM ad MG est sicut proportio ZF ad FG. Et ZH est maior quam ZF, et GH est minor quam GF. Ergo proportio ZH ad GH est maior quam proportio ZF ad FG; est ergo maior quam proportio ZM ad MG. Linea ergo que dividit angulum ZHG in duo equalia secat lineam MG; secat ergo lineam EG. Angulus igitur GHE est maior angulo EHZ. 
◉ Now let us draw line ZLH at random, let us connect EH, HG, and ZG, and let us extend FE to M [on GZ]. Accordingly, ZM:MG = ZF:FG [by Euclid, VI.3, since FM bisects angle ZFG]. Furthermore, ZH > ZF [by Euclid, III.7], and GH < GF. Hence, ZH:GH > ZF:FG, so [ZH:GH] > ZM:MG [which = ZF:FG]. Consequently, the line that bisects angle ZHG intersects line MG, so it [also] intersects line EG. Therefore, angle GHE > angle EHZ. 
◉ Et ponamus angulum EHR equalem angulo EHZ. Linea ergo HR secat lineam GF, et secat lineam EG; secet ergo lineam EG in R. Ergo due linee ZH, HR convertuntur per angulos equales, et erit L ymago R. Dico ergo quod forma cuiuslibet puncti linee GR convertitur ad visum Z ex puncto arcus FH, et non ex alio. 
◉ Let us take angle EHR = angle EHZ. Line HR therefore intersects line GF, and it [also] intersects line EG, so let it intersect line EG at R. Hence, the two lines ZH and HR are reflected at equal angles [ZHE and RHE], and L [where cathetus REL intersects line of reflection ZLH] will be the image of R [for center of sight Z]. I say, then, that the form of any point on line GR is reflected to the center of sight Z from a point on arc FH, and from no other [arc]. 
◉ Huius rei demonstratio est quoniam in capitulo de ymagine quinto tractatu, in duabus figuris viginti septem et viginti octo, dictum est quod duo arcus AB, DO non possunt esse tales quod ex illis convertetur aliquid de linea EO ad Z, et arcus BO non est de speculo. Non ergo remanet nisi arcus AD. 
◉ The proof of this is [based on] both figures 27 and 28 in chapter [2] on image [formation] in book 5, where it has been shown that the two arcs AB and DO cannot be such that anything on line EO will be reflected from them to [center of sight] Z, and the mirror does not extend to arc BO.⁑ Consequently, only arc AD is left [for the reflection]. 
◉ Sed in tricesima quinta figura dictum est quod forma cuiuslibet puncti dyametri EO convertitur ad aliquod punctum arcus AD, et in tricesima sexta capitulo de ymagine patuit quod numquam convertitur forma puncti R ad Z ex arcu AD nisi uno solo puncto. Forma ergo cuiuslibet puncti linee GR convertitur ad Z ex uno solo puncto arcus AD. 
◉ However, in the thirtyfifth proposition [of book 5] it has been shown that the form of any point on diameter EO is reflected at some point on arc AD, and in the thirtysixth [proposition] of chapter [2, book 5] on image [formation] it was demonstrated that the form of point R is reflected to Z from only one point on arc AD.⁑ Therefore, the form of any point on line GR is reflected to Z from one point only on arc AD. 
◉ Et ponamus in linea GR punctum C. Forma ergo C convertitur ad Z ex uno puncto arcus AD. Dico ergo quod illud punctum non erit nisi in arcu FH. Sin autem, convertatur forma C ad Z ex U, quod est in arcu AF, et continuemus lineas ZU, CU, GU, EU. 
◉ Let us take point C on line GR [in figure 6.7.29a, p. 136]. The form of C is therefore reflected to Z from one point on arc AD. I say, then, that that point will lie only on arc FH. For if such is not the case, let the form of C be reflected to Z from U, which lies on arc AF, and let us connect lines ZU, CU, GU, and EU. 
◉ Linea ergo GU erit maior linea GF, et ZU est minor quam ZF; ergo proportio GU ad ZU maior proportione GF ad FZ. Ergo est maior proportione GM ad MZ. Linea ergo que dividit angulum GUZ in duo equalia secat lineam ZM; secat ergo ZE. Angulus ergo GUE est minor angulo EUZ; ergo angulus CUE multo minor est angulo EUZ, et similiter de quolibet puncto arcus AU. Forma ergo C non convertitur ad Z nisi ex arcu HF. 
◉ Therefore, line GU > line GF, and ZU < ZF, so GU:ZU > GF:FZ. Hence, [GU:ZU] > GM:MZ [which = GF:FZ, by previous conclusions]. The line that bisects angle GUZ therefore intersects line ZM, so it [also] intersects ZE. Consequently, angle GUE < angle EUZ; so a fortiore angle CUE < angle EUZ, and the same holds for any [other] point on arc AU[F]. The form of C is therefore reflected to Z from arc [D]HF only. 
◉ Et dico quod non potest converti ex arcu HD. Quod si fuerit possibile, convertatur ex Q, quod est in arcu HD, et continuemus lineas ZQ, CQ, RQ, ZR, EQ, et extrahamus EH ad N. Linea ergo ZQ est maior quam ZH, et linea QR est minor quam HR; ergo proportio ZQ ad QR est maior proportione ZH ad HR, que est sicut proportio ZN ad NR. Linea ergo que dividit angulum ZQR in duo equalia secat lineam NR; secat ergo lineam ER. Angulus ergo RQE est maior angulo EQZ; angulus ergo EQC est multo maior angulo EQZ. Hoc idem sequitur in omni puncto arcus HD; forma ergo C non convertitur ad Z ex arcu HD, neque ex arcu AF. 
◉ I say, furthermore, that it cannot be reflected [to Z] from arc HD. For if that were possible, let it be reflected from Q, which lies on arc HD [in figure 6.7.29b, p. 137], and let us connect lines ZQ, CQ, RQ, ZR, and EQ, and let us extend EH to N. Therefore, line ZQ > [line] ZH, and line QR < [line] HR, so ZQ:QR > ZH:HR, which = ZN:NR [by Euclid, VI.3, since angles ZHE and RHE were constructed equal]. The line that bisects angle ZQR therefore intersects line NR, so it intersects line ER. Consequently angle RQE > angle EQZ, so a fortiore angle EQC > angle EQZ. The same result follows for any [other] point on arc HD, so the form of C is not reflected to Z from arc HD or from arc AF. 
◉ Sed iam patuit quod omnino debet converti ex arcu AD. Forma ergo C non convertitur ad Z nisi ex aliquo puncto arcus FH. Convertatur ergo ex T, et continuemus lineas CT, ET, ZT. Quia ergo T est inter duo puncta F, H, erit linea ZT inter duas lineas ZF, ZH. Linea ergo ZT secat lineam KL. Secet ergo ipsam in I. I ergo est ymago C, et C nullam habet ymaginem nisi I. 
◉ However, it has already been shown that it absolutely must be reflected from arc AD. Consequently, the form of C is only reflected to Z from some point on arc FH. Accordingly, let it be reflected from T [in figure 6.7.29c, p. 137], and let us connect lines CT, ET, and ZT. Therefore, since T lies between the two points F and H, line ZT will lie between the two lines Z[K]F and Z[L]H. Line ZT therefore intersects line KL. Let it intersect it at I, then. I is therefore the image of C, and C has no image other than I. 
◉ Et sic declarabitur quod ymago cuiuslibet puncti linee GR erit punctum linee KL. KL ergo est ymago GR, et KL est linea recta, quia est pars dyametri circuli. Et GR est linea recta, quia est etiam pars dyametri circuli. Z ergo comprehendit formam GR recte in speculo AB sperico, et hoc est quod voluimus. 
◉ And it will be demonstrated in this way that the image of any point on line GR will be a point on line KL. Thus, KL is the image of [the entire line] GR, and KL is a straight line because it is a segment of the circle’s diameter. GR is also a straight line because it too is a segment of the circle’s diameter. Thus, in [concave] spherical mirror AB, Z perceives the form of GR according to its proper [lefttoright] orientation, and this is what we wanted [to prove]. 
◉ [PROPOSITIO 30] Et iteremus formam [FIGURE 6.7.30, p. 320], et revolvamus super lineam GR a duobus lateribus duos arcus, quomodocumque sit, scilicet GNR, GQR, et sit arcus GNR non secans lineam GH. Et ponamus in linea GR punctum M, quomodocumque sit. Forma ergo M convertitur ad Z ex puncto arcus FH. Convertatur ergo ex T, et continuemus lineas ZT, MT. 
◉ [PROPOSITION 30] Now let us copy the [previous] figure, and let us circumscribe two random arcs on both sides of line GR, namely, GNR and GQR [in figure 6.7.30, p. 138], and let arc GNR not intersect line GH. Let us select point M at random on line GR. The form of M is therefore reflected to Z from [some] point on arc FH. Let it be reflected accordingly from T, and let us connect lines ZT and MT. 
◉ Duo ergo anguli ZTE, ETM sunt equales; linea ergo MT secabit arcum GNR. Secet ergo ipsum in N, et extrahamus lineam TM in parte M. Secabit ergo arcum GQR; secet ergo in puncto Q. Et continuemus NE, et extrahatur recte. Secabit ergo ZT sub lineam KL. Secet ergo illam in I. Et continuemus QE et extrahamus ipsam recte. Secabit ergo ZT super KL. Secet ergo ipsam in C. 
◉ Hence, the two angles ZTE and ETM are equal, so line MT will intersect arc GNR. Let it intersect that arc at N, then, and let us extend line TM on the side of M. It will therefore intersect arc GQR, so let it intersect at point Q. Next let us connect NE and extend it in a straight line. It will therefore intersect ZT below line KL. So let it intersect that [line] at I. Then let us connect QE and extend it in a straight line. Accordingly, it will intersect ZT above KL. Let it intersect that line at C, then. 
◉ Quia ergo duo anguli T sunt equales, erit I ymago N, et duo puncta K, L sunt ymagines duorum punctorum G, R. Ymago ergo arcus GNR est linea transiens per puncta K, I, L, ut linea KIL. Sed linea KIL est convexa ex parte visus, et arcus GNR est convexus ex parte speculi. Z ergo comprehendet formam linee GNR convexe lineam convexam. 
◉ Consequently, since the two angles [ZITE and NTE] at T are equal, I will be the image of N, and the two points K and L are the images of the two points G and R. The image of arc GNR is therefore a line passing through points K, I, and L, i.e., line KIL. But line KIL is convex with respect to the eye, and arc GNR is convex with respect to the mirror. Z will therefore perceive the form of convex line GNR [as] a convex line. 
◉ Et quia duo anguli T sunt equales, erit C etiam ymago Q, et erit linea LCK ex parte visus concava ymago arcus GQR concavi ex parte superficiei speculi. Z ergo comprehendet formam arcus GQR concavi lineam concavam. 
◉ Moreover, since the two angles [ZCTE and QTE] at T are equal, C will also be the image of Q, and line LCK, which is concave with respect to the eye, will be the image of arc GQR, which is concave with respect to the mirror’s surface. Z will therefore perceive the form of concave arc GQR [as] a concave line. 
◉ In speculis ergo concavis ex quibusdam sitibus comprehenditur linea convexa convexa, et concava concava. 
◉ Thus, in concave [spherical] mirrors a convex line is perceived [as] convex, and a concave [line as] concave in various situations. 
◉ [PROPOSITIO 31] Item sit speculum concavum in quo sit circulus ABD maximus [FIGURE 6.7.31, p. 321], et centrum G, et extrahamus lineam BG, quomodocumque sit, et dividamus ex ipsa lineam GT maiorem medietate. Et extrahamus ex T lineam ETZ perpendiculariter, et sit utraque ET, TZ equalis TG. Et continuemus ET, EG, GZ. 
◉ [PROPOSITION 31] Now let there be a concave [spherical] mirror containing great circle ABD [figure 6.7.31, p. 138], let G be the center, let us draw line BG at random, and let us cut from it line GT longer than its half. Let us then draw line ETZ from T orthogonal [to BG], and let both ET and TZ be equal to TG. Let us connect ET, EG, and GZ. 
◉ Et describamus circa triangulum EGZ circulum. Secabit ergo circulum AB in duo puncta, nam punctus T est centrum huius circuli, et TG est maior TB. Secet ergo iste circulus circulum AB in duobus punctis A, D, et continuemus lineas GA, GD, EA, EB, ED, ZA, ZB, ZD. 
◉ Let us then circumscribe a circle around triangle EGZ. It will therefore intersect circle AB at two points, for point T is the center of this [new] circle, and TG > TB [by construction]. So let this circle intersect circle AB at the two points A and D, and let us connect lines GA, GD, EA, EB, ED, ZA, ZB, and ZD. 
◉ Quia ergo due linee ET, TZ sunt equales, erunt due linee EB, BZ converse per angulos equales. Et quia duo arcus EG, GZ sunt equales, due linee EA, AZ convertuntur per angulos equales, et due linee ED, DZ convertentur per angulos equales. 
◉ Accordingly, since the two lines ET and TZ are equal [because they are both equal to TG, by construction], the two lines EB and BZ will be reflected at equal angles [i.e., EBG and GBZ subtended by equal arcs]. Also, since the two arcs EG and GZ are equal, the two lines EA and AZ are reflected at equal angles [EAG and GAZ subtended by those equal arcs], and the two lines ED and DZ will [also] be reflected at equal angles [EDG and GDZ subtended by equal arcs]. 
◉ Et quia GT est maior quam TB, erit GE maior quam EB. Angulus ergo EBG est maior angulo EGB, et angulus EGB est semirectus. Ergo duo anguli EGB, EBG simul sunt maiores recto. Ergo angulus BEG est recto minor, et angulus EGZ est rectus. Ergo due linee EB, GZ concurrent extra circulum in parte BZ. Concurrant ergo in M. 
◉ Since GT > TB [by construction], GE > EB. Hence, angle EBG > angle EGB, and angle EGB is half a right angle. The two angles EGB and EBG thus sum up to more than a right angle. Consequently, angle BEG < a right angle, whereas angle EGZ is a right angle. The two lines EB and GZ will therefore intersect outside the circle [EGZ] on the side of BZ. So let them intersect at M. 
◉ Et quia ED est intra angulum MEG, concurret cum linea GM. Concurrant ergo in L. Et quia GB transit per centrum EGZ circuli, erit portio AEG minor semicirculo. Ergo angulus AEG est obtusus, et angulus EGZ est rectus. Ergo ille due linee AE, ZG concurrent in parte EG. Concurrant ergo in F. Si ergo visus fuerit in E, et Z in aliquo visibili, tunc puncta M, L, F erunt ymagines Z. Sic ergo Z comprehenditur in tribus locis. 
◉ In addition, since ED lies within angle MEG, it will intersect line GM. Let them intersect at L. And since GB passes through the center of circle EGZ, segment AEG < a semicircle. Therefore, angle AEG is obtuse, whereas angle EGZ is right. The two lines AE and ZG will thus intersect on the side of EG. Let them intersect at F, then. Accordingly, if the center of sight is at E, and if Z lies on some visible object, points M, L, and F will be images of Z. Z is thus perceived at three places. 
◉ Item extrahamus ex E lineam ad arcum DZ, quomodocumque sit, et sit EK. Et continuemus GK, et secet arcum DZ in K, et continuemus lineas KZ, GK. Quia ergo arcus EG, GZ sunt equales, erunt duo anguli EKG, GKZ equales. Ergo angulus EKG est maior angulo GKZ. Sit ergo angulus GKN equalis angulo EKG. Due ergo linee EK, KN convertentur per angulos equales. Et extrahamus EK ad Q. Erit ergo Q ymago N respectu E. 
◉ To continue, let us draw a line at random from E to arc DZ, and let it be EK [see inset to figure 6.7.31]. Let us connect GK, let it intersect arc DZ at K, and let us connect lines KZ and GK. Therefore, since arcs EG and GZ are equal, the two angles EKG and GKZ will be equal. [Let us extend GK to point K’ on the mirror, and let us connect EK’ and ZK’]. Consequently, angle EK’G > angle GK’Z.⁑ Let angle GK’N = angle EK’G, then. Consequently, the two lines EK’ and K’N will be reflected at equal angles. Let us then extend EK’ to Q. Q will therefore be the image of N with respect to [center of sight] E. 
◉ Et ymaginemur superficiem exeuntem a linea MGF perpendiculariter super circulum ABD, et extrahamus ex Z lineam in hac superficie perpendicularem super GZ, et transeat in utramque partem. Sit ergo CZR. Et ponamus G centrum, et in longitudine GN faciamus arcum circuli CNR. Secabit ergo lineam CR in duobus punctis, et sint C, R. Et continuemus lineas GC, GR. Erunt ergo in superficie perpendiculari super superficiem ABG. Et extrahamus GC, GR recte, et super G, et in longitudine GQ faciamus arcum circuli. Secabit ergo duas lineas GC, GR. Secet in S, O. 
◉ Let us now imagine a plane passing along line MGF perpendicular to circle ABD [as represented in figure 6.7.31a, p. 139], let us draw a line from Z in this plane perpendicular to GZ, and let it extend on both sides [of Z]. Accordingly, let it be CZR. Let us then take G as a centerpoint, and let us produce arc CNR of a circle with radius GN. It will therefore intersect line CR at two points, and let them be C and R. Let us connect lines GC and GR. They will thus lie in a plane perpendicular to plane ABG [of the mirror]. Let us then extend GC and GR in a straight line, and at point G let us produce the arc of a circle with radius GQ. It will thus intersect the two lines GC and GR. Let it intersect [them] at S and O. 
◉ Quia ergo superficies circuli ABD est perpendicularis super superficiem duarum linearum GC, GR erunt duo anguli EGS, EGO recti. Erit ergo utraque superficies EGS, EGO perpendicularis super superficiem SGO, et utraque istarum superficierum facit in speculo circulum magnum comparem circulo ABD. Punctum ergo compar puncto K circuli quem facit superficies EGC, convertuntur ex ipso secundum angulos equales due linee inter duo puncta E, C. 
◉ Hence, since the plane of circle ABD [on the mirror] is perpendicular to the plane of the two lines GC and GR, the two angles EGS and EGO will be right. Both planes EGS and EGO will thus be perpendicular to plane SGO, and both of those planes cut a great circle on the mirror that is equivalent to circle ABD. From the counterpart of point K’ in the [great] circle formed by plane EGC, then, two lines between points E and C are reflected at equal angles.⁑ 
◉ Et linee GC, GR sunt equales, et linee GS, GQ, GO sunt equales, et Q est ymago N, et S est ymago C, et O est ymago R. Ymago ergo arcus CNR convexi ex parte speculi est arcus SQO concavus ex parte visus. 
◉ Moreover, lines GC and GR are equal, lines GS, GQ, and GO are equal, and Q is the image of N, S is the image of C, and O is the image of R. Therefore, the image of arc CNR, which is convex with respect to the mirror, is arc SQO, which is concave with respect to the center of sight. 
◉ Et L est ymago Z, et duo puncta S, O sunt ymagines C, R. Ymago ergo linee CZR recte est linea transiens per puncta S, L, O, et talis linea est concava ex parte visus. 
◉ Meantime, L is the image of Z, and the two points S and O are the images of C and R. Consequently, the image of straight line CZR is a line passing through points S, L, and O, and such a line is concave with respect to the center of sight. 
◉ Et signemus lineam transeuntem per puncta S, L, O, et extrahamus lineam EG ad H. Si ergo speculum non pervenit ad duo puncta B, H, sed alter duorum terminorum suorum fuerit inter duo puncta B, D, et reliquus fuerit infra H, et visus fuerit in E, et due linee RZC, RNC fuerint in aliquo visibili, tunc forma linee RZC recte erit concava, scilicet SLO, et forma arcus RNC convexi erit etiam linea concava, scilicet SQO. Et RZC recta habebit unam ymaginem, et arcus RNC habebit unam ymaginem. 
◉ Let us now draw the line passing through points S, L, and O, and let us extend line EG to H. Accordingly, if the mirror does not reach the two points B and H, but one of its two limits lies between the two points B and D, while the other lies inside of H [i.e., between H and D], and if the center of sight lies at E, while the two lines RZC and RNC lie on some visible object, the form of straight line RZC will be concave, i.e., SLO, and the form of convex arc RNC will also be a concave line, i.e., SQO. Furthermore, straight line RCZ will have a single image, and arc RNC will [also] have a single image. 
◉ Item extrahamus BG ad I, et continuemus lineas EI, IZ. Iste ergo due linee convertentur secundum angulos equales, et EI secabit FG; secet ergo in T. T ergo erit ymago Z. Puncta ergo M, L, T, F sunt ymagines Z. Et si speculum excesserit duo puncta A, I, et visus fuerit in E, et deorsum aspicientis in speculo fuerit ex parte arcus AI, comprehenderit totum arcum IDA. 
◉ Now let us extend BG to I, and let us connect lines EI and IZ. Those two lines will therefore be reflected at equal angles, and EI will intersect FG, so let it intersect at T’. Hence, T’ will be the image of Z. Points M, L, T’, and F will therefore [all] be images of Z. And if the mirror extends beyond the two points A and I, while the center of sight lies at E, and if the viewer faces the mirror on the side of arc AI, he will perceive the entire arc IDA. 
◉ Tunc Z videbitur in quattuor locis, scilicet in L, M, T, F, et videbit duo puncta R, C in duobus punctis S, O, et sic linea RZC habebit quattuor ymagines concavas. Una transit per puncta S, M, O, scilicet linea SMO; secunda pertransibit per puncta S, L, O, scilicet linea SLO; tertia transibit per puncta S, T, O, scilicet linea STO; quarta transibit per puncta S, F, O, scilicet linea SFO. 
◉ Consequently, Z will appear at four places, i.e., at L, M, T’, and F, and he will see the two points R and C at the two points S and O, so line RZC will have four concave images. One passes through points S, M, and O, i.e., line SMO; a second will pass through points S, L, and O, i.e., line SLO; a third will pass through points S, T’, and O, i.e., line ST’O; and a fourth will pass through points S, F, and O, i.e., line SFO. 
◉ Patet ergo ex hac figura quod linea recta in speculis concavis comprehenditur concava, et convexa comprehenditur concava, et quod recta habet plures formas concavas. 
◉ From this proposition it is therefore clear that in concave [spherical] mirrors a straight line is perceived as concave, a convex [one] is also perceived as concave, and a straight [line] has several concave forms.⁑ 
◉ [PROPOSITIO 32] Item sit speculum concavum [FIGURE 6.7.32, p. 322] per cuius centrum transeat superficies, et faciat circulum ABG, et sit centrum D. Et extrahamus ex D lineam, quomodocumque sit, et sit DG, et transeat extra circulum. Et extrahamus ex D in superficie huius circuli lineam perpendicularem super lineam DG, et sit DA. Et abscindamus de angulo ADG recto particulam parvam, quomodocumque sit, et sit angulus GDE, ita quod inter angulum rectum et angulum ADE sit multiplum anguli EDG, et dividamus angulum ADE in duo equalia per lineam DB. Et abscindamus distinctionem equalem angulo EDG, et extrahamus ex D lineam continentem cum DB angulum rectum, et sit DT. 
◉ [PROPOSITION 32] Now let there be a concave [spherical] mirror through whose center a plane passes, let it produce [great] circle ABG [in figure 6.7.32, p. 140], and let D be its center. From D let us draw a line at random, let it be DG, and let it extend beyond the circle. From point D let us extend a line in the plane of the circle perpendicular to line DG, and let it be DA. Let us then cut from right angle ADG a small subangle GDE at random such that the difference between right angle [ADG] and angle ADE consists of several [increments of] angle EDG,⁑ and let us bisect angle ADE with line DB. Let us also cut [from angle ADG] a subangle [ADZ] equal to angle EDG, and let us extend a line from D that forms a right angle with DB [i.e., BDT], and let it be DT. 
◉ Et extrahamus AD in parte D, et sit DK, et extrahamus ex Z lineam continentem cum ZD angulum equalem angulo KDT. Hec ergo linea concurret cum DA, nam duo anguli KDT, ADZ sunt minores duobus rectis. Concurrant ergo in H. Angulus ergo ZHD est equalis angulo ZDT. 
◉ Now let us extend AD on the side of D, let it form DK, and let us extend a line [ZH] from Z that forms with ZD an angle [DZH] equal to angle KDT. This line will therefore intersect DA because the two angles KDT [which = DZH, by construction] and ADZ sum up to less than two right angles. So let them intersect at H. Angle ZHD is thus equal to angle ZDT.⁑ 
◉ Et extrahamus ex Z lineam continentem cum ZH angulum equalem angulo BDK obtuso, et sit ZL. Duo ergo anguli LZD, BDZ sunt minores duobus rectis; linea ergo ZL concurret cum DB. Concurrat ergo in L. 
◉ Then from Z let us extend line ZL to form an angle [HZL] with ZH equal to obtuse angle BDK. Accordingly, the two angles LZD and BDZ sum up to less than two right angles, so line ZL will intersect [line] DB.⁑ Let it therefore intersect at L. 
◉ Et continuemus LH, et circa triangulum HLD faciamus circulum DHL. Transibit ergo per Z, quia duo anguli LZH, LDH sunt equales duobus rectis. Anguli ergo LHZ, LDZ sunt equales, quia basis eorum est idem arcus. Sed angulus ZHD est equalis angulo ZDT; remanet ergo angulus LHD equalis angulo LDT. Et angulus LDT est rectus; ergo angulus LHD est rectus. 
◉ Let us then connect LH, and let us form circle DHL around triangle HLD. It will therefore pass through Z because the two angles LZH and LDH sum up to two right [angles, by Euclid, III.22]. Consequently angles LHZ and LDZ are equal because they are subtended by the same arc [LZ]. But angle ZHD = angle ZDT [by previous conclusions], so it follows that angle LHD = angle LDT. But angle LDT is right [by construction], so angle LHD is right. 
◉ Et abscindamus ex linea DE lineam DM equalem DH, et continuemus LM. Angulus ergo LMD est rectus; circulus ergo LHD transit per M et secat arcum HE in puncto compari Z. Secet ergo in F, et continuemus DF. Angulus ergo LDF erit equalis angulo LDZ, quia arcus LM est equalis arcui LH, et arcus MF est equalis arcui ZH. Ergo arcus FMD est equalis arcui ZHD. 
◉ Now from line DE let us cut line DM equal to [line] DH, and let us connect LM. Angle LMD is therefore right, so circle LHD passes through M and cuts arc HE at a point equivalent to Z.⁑ Accordingly, let it intersect at F, and let us draw DF. Angle LDF will therefore be equal to angle LDZ because arc LM = arc LH [by construction according to the bisection of angle ADE by DB], and arc MF = arc ZH [so arc FL subtending angle FDL = arc LZ subtending angle LDZ]. Consequently, arc FMD = arc ZHD. 
◉ Et continuemus lineas HB, HF, FM, FZ, FB. Angulus ergo BHD erit acutus, et angulus GDH erit rectus. Ergo linea HB concurret cum linea DG extra circulum. Concurrant ergo in Q. HF ergo concurret etiam cum DG extra circulum; concurrant ergo in N. 
◉ Let us now draw lines HB, HF, FM, FZ, and FB. Angle BHD will thus be acute, while angle GDH will be right [by construction]. Therefore, line HB will intersect line DG outside the circle. Let them intersect at Q, then. HF will therefore also intersect DG outside the circle, so let them intersect at N. 
◉ Et extrahamus FB quousque secet arcum LZ. Secet ergo in R, et continuemus RM. Angulus ergo FRM, qui est in circumferentia, respicit arcum FM, et angulus FBM est maior angulo FRM, et angulus FBM est in circumferentia ABG. Ergo si BM linea extrahatur in parte M, abscindet de circulo ABG arcum maiorem simili arcus FM. 
◉ Let us then extend FB until it intersects arc LZ. Accordingly, let it intersect at R, and let us connect RM. Angle FRM, which lies on the circumference [of circle ZDF], is thus subtended by arc FM, and angle FBM > angle FRM, but angle FBM lies on the circumference of [circle] ABG. Therefore, if line BM is extended on the side of M, it will cut a larger arc on circle ABG than the counterpart FM [it cuts on circle ZDF]. 
◉ Et arcus FM est duplus similis arcus FE. Et arcus FE est equalis arcui ZA, et arcus ZA est equalis arcui EG, et arcus FE est equalis arcui EG. Ergo arcus GF est duplum arcus GE; ergo arcus GF est similis arcui FM. 
◉ But arc FM [in circle ZDF] is twice its counterpart FE [in circle ABG].⁑ Moreover, arc FE = arc ZA [because arcs FM and ZH are equal], whereas arc ZA = arc EG [by construction], and [so] arc FE = arc EG. Consequently, arc GF is twice arc GE, so arc GF [in circle ABG] is the equivalent [in degrees of arc] of arc FM [in circle FDZ]. 
◉ Si ergo BM extrahatur recte in partem M, abscindet de circulo ABG arcum ultra punctum F maiorem arcu FG. Linea ergo BM secabit lineam DG inter duo puncta G, D. Secet ergo in O. Et extrahamus lineam FM, et secet DO in U; et extrahamus BM in parte B, et secet arcum LR in C. Et continuemus CD. 
◉ Hence, if BM is extended in a straight line on the side of M, it will cut an arc on circle ABG beyond point F that is greater than arc FG. Line BM will therefore intersect line DG between the two points G and D. Accordingly, let it intersect at O. Let us then extend line FM, and let it intersect DO at U; let us also extend BM on the side of B, and let it intersect arc LR at C. Let us then connect CD. 
◉ Quia ergo angulus BFZ est in circumferentia ABG, erit angulus BFZ dimidium anguli BDZ. Sed angulus BDZ est multiplus anguli ZDA; ergo angulus RFZ est multiplus anguli ZDH. Ergo arcus RZ est multiplus arcus ZH, et arcus CZ est maior arcu RZ; ergo arcus CZ est multiplus arcus ZH. 
◉ Accordingly, because angle BFZ lies on the circumference of [circle] ABG, angle BFZ will be half of angle BDZ [by Euclid, VI.33]. But angle BDZ is several times larger than angle ZDA [by construction], so angle R[B]FZ is several times larger than angle ZDH[A]. Consequently, arc RZ is several times the size of arc ZH, and arc CZ > arc RZ, so arc CZ is several times the size of arc ZH. 
◉ Et continuemus CH. Angulus ergo CHD cum angulo CMD sunt equales duobus rectis; ergo angulus CHD est equalis angulo BME. Ergo angulus ZHD addit super angulum CHD angulum CHZ, qui est equalis angulo CDZ, et angulus CDZ est multiplus anguli ZDA. Ergo angulus CHZ est multiplus anguli EDG; ergo angulus ZHD excedit angulum CHD multiplo anguli EDG. Angulus ergo ZHD est equalis angulo FMD, quia arcus FMD est equalis arcui ZHD. 
◉ Let us now connect CH. Accordingly, angle CHD + angle C[B]MD = two right angles [by Euclid, III.22], so angle CHD = angle BME [adjacent to BMD]. Therefore, angle ZHD = angle CHD + angle CHZ, which = angle CDZ [since they are both subtended by arc CZ in circle ZDF], and angle CDZ is several times larger than angle ZDA. Therefore, angle CHZ is several times larger than angle EDG [which = angle ZDA, by construction], so angle ZHD exceeds angle CHD by several [increments of] angle EDG. Hence, angle ZHD = angle FMD because arc FMD = arc ZHD [by previous conclusions]. 
◉ Et angulus CHD, ut declaravimus, est equalis angulo BME. Ergo angulus FMD excedit angulum BME multiplo anguli EDG. Ergo angulus FMD excedit angulum OMD multiplo anguli EDG. Et MOG angulus excedit angulum OMD angulo EGD; ergo angulus FMD excedit angulum MOG multiplo anguli EDG. 
◉ Moreover, as we demonstrated [just above], angle CHD = angle BME. Angle FMD thus exceeds angle BME by several [increments of] angle EDG. Therefore, angle FMD exceeds angle OMD by several [increments of] angle EDG [because angle OMD = vertical angle BME]. But angle MOG exceeds angle OMD by [one increment of] angle EDG [by Euclid, I.32], so angle FMD exceeds angle MOG by several [increments of] angle EDG. 
◉ Et angulus FMD excedit angulum MUD angulo EDG solo. Ergo angulus MUD est maior angulo MOG; ergo angulus MOU est maior angulo MUO. Ergo linea MU est maior linea MO. Et quia arcus ZHD est equalis arcui FMD, erunt duo anguli HFD, MFD equales. Due ergo linee HF, FU convertentur equaliter, et similiter HB, BO convertentur equaliter. Q ergo est ymago O, et N est ymago U. 
◉ In addition, angle FMD exceeds angle MUD by only [one increment of] angle EDG [by Euclid, I.32]. Therefore, angle MUD > angle MOG, so angle MOU [adjacent to MOG] > angle MUO [adjacent to MUD]. Hence, line MU > line MO [since MU subtends the larger angle]. And since arc ZHD = arc FMD [by previous conclusions], the two angles HFD and MFD will be equal. The two lines HF and FU will therefore reflect at equal [angles], and likewise HB and BO will reflect at equal [angles]. Consequently, Q is the image of O, and N is the image of U [from the perspective of H, as the center of sight]. 
◉ Et extrahamus ex M lineam equidistantem linee HQ, et sit MS, et extrahamus ex M etiam lineam equidistantem linee HN, et sit MP. Quia ergo angulus HND est maior angulo HQD, erit angulus MPO maior angulo MSO. P ergo erit inter duo puncta S, U, et quia angulus HDN est rectus, erit angulus HND acutus. Ergo angulus MPD est acutus; ergo angulus MPS est obtusus. Ergo linea MS est maior quam MP. 
◉ From M let us then extend a line parallel to line HQ, let it be MS, and let us also extend a line from M parallel to line HN, and let it be MP. Thus, since angle HND > angle HQD, angle MPO > angle MSO. P therefore lies between the two points S and U, and because angle HDN is right [by construction], angle HND will be acute. Accordingly, angle MPD is acute, so angle MPS [adjacent to it] is obtuse. Line MS is therefore longer than [line] MP. 
◉ Sed MU est maior quam MO, ut diximus; ergo proportio SM ad MO est maior quam proportio PM ad MU, et proportio SM ad MO est sicut proportio QB ad BO, quia MS est equidistans BQ. Et similiter proportio PM ad MU est sicut proportio NF ad FU; ergo proportio QB ad BO est maior quam proportio NF ad FU. Et proportio QB ad BO est sicut proportio QD ad DO, et proportio NF ad FU est sicut proportio ND ad DU, ut declaravimus in vicesima quinta figura capitulo de ymagine. Ergo proportio QD ad DO est maior quam proportio ND ad DU. 
◉ But MU > MO, as we [just] established, so SM:MO > PM:MU, and SM:MO = QB:BO because MS is parallel to BQ. Likewise, PM:MU = NF:FU, so QB:BO > NF:FU. Moreover, QB:BO = QD:DO, whereas NF:FU = ND:DU, as we showed in the twentyfifth proposition of chapter [2, book 5] on image [formation].⁑ Therefore, QD:DO > ND:DU. 
◉ Hiis preostensis, iteremus circulum, et perficiamus demonstrationem, ne multiplicentur linee et dubitentur littere. Sit ergo circulus in secunda forma ABG [FIGURE 6.7.32a, p. 323], et centrum D, et extrahamus lineam DQ. Et sit DU equalis DU in prima forma, et DO equalis DO in prima forma. Et DQ sit compar sibi in prima forma, et similiter DN. 
◉ Now that these points have been established, let us redraw the circle and finish the proof so as to avoid adding lines and confusing letterdesignations. Accordingly, let ABG [in figure 6.7.32b, p. 142] be the circle in the second version [of figure 6.7.32, p. 140], let D be its center, and let us draw line DQ. Let DU [in the second version] be equivalent to DU in the original version, and let DO [in the second version] be equivalent to DO in the original version. Also, let DQ [in the second version] be equivalent to its counterpart in the original version, and likewise for DN. 
◉ Et extrahamus super DQ DH perpendicularem super superficiem circuli, et sit DH equalis sibi in prima forma. Angulus ergo HDQ erit rectus, et circulus quem facit HDQ in speculo erit ex circulis ex quibus forma convertitur. Et erit arcus quem mensurant linee HD, DQ equalis arcui AG in primo circulo. Et ex duobus punctis istius comparibus duobus punctis B, F convertentur linee ad duo puncta U, O equaliter. Erit ergo Q ymago O, et N ymago U. 
◉ Let us then draw DH’ perpendicular to DQ [as well as] to the plane of the circle, and let DH’ be equivalent to its counterpart [DH] in the original version. Angle H’DQ will therefore be right, and the [great] circle that [the plane containing] H’DQ produces in the mirror will be among the circles [like ABG] within which a form is reflected. Furthermore, the arc that lines H’D and DQ measure off [in the great circle produced on the mirror by plane H’DQ] will be equal to arc AG in the original circle. And from the two points on this [arc] that are equivalent to the two points B and F [on arc AB in the original circle] lines from the two points U and O will be reflected at equal angles [to point H’]. Q will therefore be the image of O [for center of sight H’], and N [will be] the image of U.⁑ 
◉ Et extrahamus ex U perpendicularem lineam in superficie circuli ABG super lineam DU, et sit ZUE. Et sit D centrum, et in longitudine DO faciamus arcum circuli. Secabit ergo lineam ZUE in duobus punctis. Secet ergo in Z, E, et sit arcus ZOE. Et continuemus DZ, DE et extrahantur extra circulum. Et circa D et in longitudine DQ faciamus arcum TQK. Secabit ergo duas lineas DZ, DE in T, K. Et continuemus TK. Secabit ergo lineam DQ in L. 
◉ From U let us produce a line perpendicular to line DU within the plane of circle ABG, and let it be ZUE. Let D be the center, and [from it] let us produce the arc of a circle with a radius of DO. It will therefore intersect line ZUE at two points. Accordingly, let it intersect at Z and E, and let it form arc ZOE. Let us then draw DZ and DE, and extend them beyond the circle. At a radius of DQ, let us produce arc TQK around D. It will therefore intersect the [extension of the] two lines DZ and DE at T and K. Let us then draw TK. Accordingly, it will intersect line DQ at L. 
◉ Quia ergo HD est perpendicularis super superficiem circuli, uterque angulus HDT, HDK erit rectus. Et utraque superficies HDT, HDK facit in superficie speculi circulum, et arcus illius qui est inter duas lineas HD, DT erit equalis arcui qui est inter HD, DQ, et similiter arcus qui est inter duas lineas HD, DK. Et utraque linea DZ, DE est equalis linee DO. Ergo hii duo arcus sunt huiusmodi quod ex illis convertentur linee secundum angulos equales ad duo puncta Z, E. Et due linee DT, DK sunt equales linee DQ; ergo punctum T est ymago Z, et K est ymago E. 
◉ Consequently, since H’D is perpendicular to the plane of the circle, both angles H’DT and H’DK will be right. Moreover, both planes H’DT and H’DK produce a [great] circle on the mirror’s surface, and the arc on it that lies between the two lines H’D and DT will be equal to the arc lying between HD and DQ [in the original figure—i.e., 6.7.32], and the same holds for the arc between the two lines H’D and DK. In addition, both lines DZ and DE are equal to line DO. These two arcs [cut on the mirror’s surface by planes H’DZ and H’DE] are therefore of the kind from which lines will be reflected at equal angles to the two points Z and E.⁑ Furthermore, the two lines DT and DK are equal to line DQ, so point T is the image of Z, and [point] K is the image of E. 
◉ Et quia linee DT, DQ, DK sunt equales, et linee DZ, DO, DE sunt equales, erit proportio DT ad DZ sicut proportio QD ad DO, et sicut proportio KD ad DE. Sed proportio QD ad DO, ut in prima figura preostendimus, est maior proportione ND ad DU. Ergo proportio DT ad DZ est maior proportione ND ad DU, et similiter proportio KD ad DE. 
◉ Since, moreover, lines DT, DQ, and DK are equal, and since lines DZ, DO, and DE are equal, DT:DZ = QD:DO = KD:DE. But, as we showed in the previous theorem [i.e., in paragraph 7.97 keyed to figure 6.7.32], QD:DO > ND:DU. Therefore, DT:DZ > ND:DU, and the same holds for KD:DE. 
◉ Et quia due linee ZD, DE sunt equales, et due linee DT, DK sunt equales, erit linea TK equidistans linee ZE. Ergo utraque proportio DT ad DZ et KD ad DE erit sicut proportio LD ad DU. Ergo proportio LD ad DU est maior proportione ND ad DU; ergo linea LD est maior linea ND. Ergo N est inter L, U. Sed N est ymago U, et duo puncta T, K sunt ymagines Z, E. Ergo ymago linee ZUE recte est linea transiens per puncta T, N, K. Et linea que transit per hec puncta est convexa, ex quibus patet quod linea recta in speculis concavis quandoque videtur convexa in quibusdam sitibus. 
◉ In addition, since the two lines ZD and DE are equal, and since the two lines DT and DK are equal, line TK will be parallel to line ZE. Therefore, DT:DZ and KD:DE will [both] be as LD:DU. Hence, LD:DU > ND:DU, so line LD > line ND. N therefore lies between L and U. But N is the image of U, and the two points T and K are the images of Z and E. As a result, the image of straight line ZUE is the line that passes through points T, N, and K. But the line that passes through these points is convex, from which it is clear that in concave [spherical] mirrors a straight line sometimes appears convex in certain situations. 
◉ Item ponamus in linea ZU punctum M, quomodocumque sit, et circa centrum M et in longitudine MU, faciamus arcum RUF. Iste ergo arcus secabit arcum UOE in duobus punctis. Secet ergo in R, F, et continuemus lineas DR, DF, et transeant recte usquoque concurrant in arcu TQK in C, I. Superficies ergo duarum linearum HD, DC faciet in speculo circulum a cuius circumferentia convertentur equaliter linee ad R, et similiter superficies duarum linearum HD, DI faciet in speculo circulum a cuius circumferentia convertentur linee ad F. C ergo est ymago R, et I est ymago F, et N est ymago U. 
◉ Now let us take some point M at random on line ZU [in figure 6.7.32d, p. 143], and around M as center let us produce arc RUF with radius MU. This arc will therefore intersect arc ZOE⁑ in two points. Accordingly, let it intersect at R and F, let us draw lines DR and DF, and let them pass in a straight line until they intersect arc TQK at C and I. The plane containing the two lines H’D and DC will therefore produce a [great] circle on the mirror from whose circumference lines from R will be reflected at equal angles, and by the same token the plane containing the two lines H’D and DI will produce a [great] circle on the mirror from whose circumference lines will be reflected to F [at equal angles]. C is therefore the image of R, I is the image of F, and N is the image of U.⁑ 
◉ Ymago ergo arcus RUF est linea transiens per C, N, I, sed hec linea erit convexa, et arcus RUF est concavus ex parte superficiei speculi. Cum ergo visus fuerit in H et unaqueque linea ZUE, ZOE, RUF fuerit in aliquo visibili, tunc linea ZUE recta comprehendetur convexa, et linea ZOE convexa comprehenditur concava, et concava convexa. Si ergo unaqueque linea ZUE, ZOE, RUF habuerit unam ymaginem, tunc forma illarum linearum erit eodem modo quo declaravimus. Et si habuerit alias ymagines, forte erunt similes aliis ymaginibus, et forte diverse. 
◉ Consequently, the image of arc RUF is the line passing through C, N, and I, but this line will be convex, whereas arc RUF is concave with respect to the mirror’s surface. Therefore, when the center of sight is at H’, and when any of the lines ZUE, ZOE, or RUF lies on some visible object, straight line ZUE will be perceived [as] convex, convex line ZOE will be perceived [as] concave, and concave [line RUF is perceived as] convex. Consequently, if each of the lines ZUE, ZOE, and RUF has [only] one image, the form of those lines will be just as we showed. But if they have additional images, they may be similar to the other images [i.e., the ones just discussed], or they may be different. 
◉ Patet ergo ex istis figuris quod linee recte in speculis concavis quandoque comprehenduntur recte, quandoque convexe, quandoque concave. Et linee convexe quandoque comprehenduntur convexe, quandoque concave, et concave quandoque comprehenduntur convexe quandoque concave. 
◉ From these propositions [i.e., 2932] it is therefore evident that straight lines are sometimes perceived [as] straight in concave [spherical] mirrors, sometimes [as] convex, and sometimes [as] concave. In addition convex lines are sometimes perceived [as] convex [and] sometimes [as] concave, and concave [lines] are sometimes perceived [as] convex [and] sometimes [as] concave. 
◉ Forme ergo superficierum visibilium comprehenduntur aliter quam sint in huiusmodi speculis, nam linee recte non sunt nisi in superficiebus rectis, et cum linea recta que existit in superficie plana comprehendatur convexa aut concava, tunc superficies in qua est comprehendetur convexa aut concava. Cum ergo visus comprehendit lineas convexas et concavas et rectas aliter quam sint, comprehendet superficies in quibus sunt aliter quam sint. 
◉ So the forms of visible surfaces are perceived [as] other than they [actually] are in these sorts of mirrors, for straight lines exist only in flat surfaces, and when a straight line that exists in a plane surface is perceived [as] convex or concave, the surface in which it lies will be perceived [as] convex or concave. Accordingly, when the eye perceives convex, concave, and straight lines other than they [actually] are, it will perceive the surfaces in which they lie other than they are. 
◉ Patet ergo ex predictis quod in omnibus que in speculis concavis comprehenduntur accidit fallacia, sed in quibusdam accidit semper et in omni positione, in quibusdam vero accidit in aliqua positione. Fallacie autem composite accidunt in hiis speculis eo modo quo in compositis, et hoc voluimus declarare. 
◉ From the foregoing, then, it is clear that in everything that is perceived in concave [spherical] mirrors, a misperception occurs, but in certain cases it occurs in every situation, without exception, whereas in certain [cases] it occurs in a specific situation. Moreover, compound misperceptions arise in these mirrors just as in the case of compound illusions [in the other mirrors],⁑ and this [is what] what we wanted to demonstrate. 
◉Capitulum octavum 
◉Chapter Eight 
De fallaciis que accidunt in speculis columpnaribus concavis 

◉ In hiis enim accidunt similia eis que accidunt in spericis concavis, accidunt enim fallacie que proveniunt ex conversione, scilicet debilitas lucis et coloris et diversitas situs et remotionis que accidunt omnibus speculis. Accidit autem in eis ex diversitate quantitatis simile illi quod accidit in speculis spericis concavis. Et videtur etiam unum visibile unum, et duo, et tria, et quattuor, et rectum et conversum secundum diversos situs, et planum videtur concavum et convexum. Ostendamus ergo qualiter in hiis speculis diversatur quantitas et numerus rei vise, et qualiter apparet rectum et conversum eo modo quo in speculis spericis concavis declaravimus. 
◉ In these [sorts of mirrors] the same things happen as happen in concave spherical [mirrors], for the misperceptions that arise from reflection [by itself] occur [in these mirrors], i.e., the weakening of light and color and the variation in situation and distance that occur in all mirrors. Moreover, variation in size occurs in these mirrors in the same way as it happens in concave spherical mirrors. Also, one visible object appears [as] one, or [as] two, or [as] three, or [as] four, and [it appears] properly oriented or reversed in various circumstances, and a flat object appears concave or convex. So let us show how the size and number of a visible object may vary in these mirrors, as well as how it appears properly oriented or reversed in the way that we demonstrated [these phenomena] in spherical concave mirrors. 
◉ [PROPOSITIO 33] Iteremus ergo primam figuram ex duabus figuris premissis in fallaciis speculorum columpnalium convexorum, et eisdem litteris. In illa autem figura [FIGURE 6.8.33, p. 324] patuit quod linee EG, GT, EB, QB, EA, AH convertuntur secundum angulos equales; et quod linee EO, HA, BQ, TG coniunguntur in O; et quod linea ABG est linea recta extensa in longitudine speculi; et quod linee GZ, BL, AD sunt perpendiculares super superficiem contingentem superficiem que transit per lineam ABG; et quod linea ABG est perpendicularis super superficiem in quo est triangulus EBO; et quod linea TQ est equalis QH, et AB equalis BG; et quod S, C, I sunt ymagines H, Q, T; et quod C est propinquius puncto E quam linea SI; et quod linea SI est in superficie trianguli UHT; et quod due linee UH, UT sunt equales; et quod due linee US, UI sunt equales; et quod due linee ES, EI sunt equales. 
◉ [PROPOSITION 33] Let us recapitulate the first of the two diagrams provided in [the analysis of] misperceptions [occurring in] convex cylindrical mirrors [i.e., figure 6.5.18, p. 127, redrawn as figure 6.8.33, p. 144], and [let us use] the same letters. Now in that proposition [i.e., proposition 18, in conjunction with proposition 17, pp. 190193 above] it was shown: that lines EG and GT, EB and QB, and EA and AH are reflected at equal angles; that lines EO, HA, BQ, and TG intersect at O;⁑ that line ABG is a straight line extended along the longitude of the mirror; that lines GZ, BL, and AD are perpendicular to the plane tangent to the [mirror’s] surface and passing along line ABG; that line ABG is perpendicular to the plane containing triangle EBO; that line TQ = [line] QH, and [line] AB = [line] BG; that S, C, and I are the images of H, Q, and T; that C lies nearer point E than [straight] line SI; that [straight] line SI lies in the plane of triangle UHT; that the two lines UH and UT are equal; that the two lines US and UI are equal; and that the two lines ES and EI are equal. 
◉ Et continuemus CU, et secet SI in F. Dividet ergo ipsam in duo equalia, nam HT est divisa in duo equalia in Q, et erit CU in superficie trianguli CUE, qui est superficies circuli B equidistantis basi speculi. Q ergo erit in superficie trianguli CUE, et C est in triangulo CEI. Ergo C est in differentia communi hiis duabus superficiebus. Sed hec differentia est linea EB; ergo C est in rectitudine EB. 
◉ Let us draw CU, and let it intersect SI at F. It will therefore bisect this line [i.e., SI] because HT is bisected at Q [by construction], and CU will lie in the plane of triangle CUE, which lies in the plane of the circle [passing through] B parallel to the base of the mirror.⁑ Q will therefore lie in the plane of triangle CUE, and C lies in triangle CEI. Hence, C lies on the common section of these two planes. But this [common] section is line EB; so C lies on a straight line [with] EB.⁑ 
◉ Et due linee HU, TU sunt sub duobus punctis D, Z, nam due linee HU, TU sunt perpendiculares exeuntes ex H, T super duas lineas contingentes duas portiones in quarum circumferentia sunt puncta A, G. Superficies ergo trianguli UHT est sub axe DLZ. 
◉ Moreover, the two lines HU and TU [in figure 6.8.33] lie outside the two points D and Z [on the axis], for the two lines HU and TU are the normals passing from H and T to the two lines that are tangent to two sections [on the cylinder’s surface] on whose periphery points A and G lie [i.e., the two elliptical sections formed on the mirror’s surfaces by planes of reflection TGE and HAE]. Consequently, the plane of triangle UHT lies outside axis DLZ. 
◉ Sed nullum punctum huius axis, quamvis exeat in infinitum, erit in superficie trianguli UHT, nam si esset, tunc, si continuaretur cum aliquo puncto linee HT linea recta, illa superficies in qua esset illa linea recta et linea HT esset tunc superficies trianguli UHT, et illa superficies esset illa in qua sunt due linee equidistantes HT, DZ. Et sic superficies in qua sunt due linee HT, DZ est superficies trianguli HUT, et sic axis erit in superficie trianguli HUT. 
◉ Even if the axis is extended to infinity, however, no point on it will lie in the plane of triangle UHT, for if it did, then if it were to be connected in a straight line with some point on line HT, the plane in which that straight line and line HT lay would be the plane of triangle UHT, and that plane would be the one in which the two parallel lines HT and DZ lie. Hence, the plane containing the two lines HT and DZ [supposedly] forms the plane of triangle HUT, so the axis will lie in the plane of triangle HUT. 
◉ Sed axis est equidistans linee HT positione, et axis secat duas lineas HU, TU. Et linea TH est in superficie trianguli UEH, que est superficies conversionis, et superficies communis huic superficiei et superficiei columpne est aliquis sector. Superficies ergo EUH secat axem columpne in uno puncto, scilicet in D, ut preostendimus. Et cum axis secet lineam HU, punctus sectionis cum linea HU erit in superficie trianguli UEH. Sed in hac superficie non est punctum per quod axis transeat preter quam D. Ergo linea HU secat axem in D. Et iam ostendimus quod HU secat eum in puncto sub D, quod est impossibile. 
◉ But the axis is parallel to line HT by construction, and the axis [supposedly] intersects the two lines HU and TU. Moreover, line TH lies in [i.e., passes through] the plane of triangle UEH, which is the plane of reflection [for objectpoint H and reflectionpoint A], and the common section of this plane and the surface of the mirror is some [elliptical] section. Therefore, plane EUH intersects the axis of the cylinder in one point, that is, D, as we showed before [in proposition 18]. And if the axis intersects line HU, the point of intersection with line HU will lie in the plane of triangle UEH. But there is no point in this plane other than D through which the axis might pass. Therefore, line HU intersects the axis at D. But it has already been shown that HU intersects it at a point beyond D, which is impossible.⁑ 
◉ Ergo axis DZ est extra superficiem UHT et propinquior puncto E quam superficies UHT. Superficies ergo in qua sunt linee HT, DZ est propinquior puncto E quam superficies UHT. Et C est in superficie in qua sunt HT, DZ, quia est in linea QL, et QL est in superficie in qua sunt HT, DZ. Ergo C est propinquior puncto E quam S, I. Sed C est in rectitudine EB. Si ergo EB exiverit in parte B, perveniet ad C; perveniat ergo ad C. 
◉ Consequently, axis DZ lies outside the plane of UHT and nearer to point E than plane UHT [i.e., between E and plane UHT]. The plane in which lines HT and DZ lie is therefore nearer to point E than plane UHT. Moreover, C lies in the same plane as HT and DZ because it lies on line QL, and QL lies in the same plane as HT and DZ. Therefore, C lies nearer to point E than [do] S and I. But C lies in a straight line with EB. If, therefore, EB is extended toward B, it will reach C, so let it reach C. 
◉ Hiis preostensis, dico quod linea SI, que est equidistans axi speculi, cum fuerit in aliquo visibili, et visus fuerit in O ex parte concavitatis columpne, et superficies speculata fuerit superficies concava, tunc SI comprehendetur ex O in speculo ABG concavo, et diversabuntur ymagines eius secundum diversitatem sue distantie ab axe. 
◉ Now that these points are established, I say that if line SI, which is parallel to the mirror’s axis, lies on some visible object, if the center of sight lies at O on the concave side of the cylinder, and if the reflecting surface is a concave surface, then SI will be perceived by O in concave mirror ABG, and its images will vary according to how its distance from the axis varies. 
◉ Cuius demonstratio est quod angulus EBM est acutus; ergo angulus LBC est acutus. Et linea EBC est in superficie circuli B, et LB est dyameter huius circuli. Ergo EB secat circulum; ergo CB est intra concavitatem speculi. 
◉ The proof of this [claim] lies in the fact that angle EBM is acute, so [vertical] angle LBC is acute. Moreover, line EBC lies in the plane of the circle [passing through] B, and LB is [on] the diameter of this circle. Hence, EB intersects the circle, so CB lies inside the mirror’s concavity. 
◉ Et similiter OB est intra concavitatem speculi, quia angulus OBL est acutus, et duo anguli OBL, CBL sunt equales, nam sunt equales duobus angulis EBM, QBM, et LB est perpendicularis super superficiem contingentem columpnam que transit per B. Forma ergo C extenditur per CB et pervenit ad B, et convertitur per BO et comprehenditur a visu ex O. 
◉ By the same token, OB lies inside the mirror’s concavity because angle OBL is acute, and the two angles OBL and CBL are equal, since they are equal to the two angles EBM and QBM, while LB is perpendicular to the plane that passes through B tangent to the cylinder. The form of C thus passes along CB and reaches B, and it is reflected along BO and perceived by the center of sight at O. 
◉ Item in quinto capitulo, cum fuimus locuti de speculis columpnaribus convexis, declaravimus quod superficies contingens columpnam in G erit sub E. Ergo EG secat superficiem contingentem; secat ergo lineam contingentem circumferentiam sectoris in G. Secat ergo sectorem et cadit intra ipsum; cadet ergo intra concavitatem speculi. Ergo due linee OG, GI sunt intra concavitatem speculi, et ZG est perpendicularis super superficiem contingentem columpnam in G, et duo anguli OGZ, IGZ sunt equales. Ergo forma I extenditur per IG et pervenit ad G, et convertitur per GO et comprehenditur ex O per lineam GO. Et similiter S extenditur per SA et convertitur per AO. 
◉ Furthermore, when we discussed convex cylindrical mirrors in chapter 5 [of book 4, paragraph 5.18, in Smith, Alhacen on the Principles, 332], we showed that the plane tangent to the cylinder at G will lie below E. Therefore, EG intersects the tangent plane, so it intersects the line tangent to G on the periphery of the [elliptical] section [formed on the mirror by the plane of reflection]. As a result, it intersects the [elliptical] section and falls inside it; so it will fall inside the concavity of the mirror. The two lines OG and GI thus lie inside the concavity of the mirror, whereas ZG is perpendicular to the plane tangent to the cylinder at G, and the two angles OGZ and IGZ are equal. Hence, the form of I passes along IG, reaches G, is reflected along GO, and is perceived at O along line GO. So too, [the form of] S passes along SA and is reflected along AO. 
◉ Et iam declaravimus, cum tractavimus de fallaciis speculorum columpnalium convexorum, quoniam due linee HU, TU sunt perpendiculares super duas superficies contingentes sectores transeuntes per duo puncta A, G. Ymago ergo S est in linea HU. Et OA linea radialis que extenditur ex visu ad punctum conversionis; ergo ymago S est in OA. H ergo est ymago S, et sic patet quod T est ymago I. 
◉ But when we dealt with misperceptions [arising] from convex cylindrical mirrors, we demonstrated [in proposition 16, lemma 5] that the two lines HU and TU are normal to the two planes tangent to the [elliptical] sections passing through the two points A and G. Therefore, the image of S lies on line HU. Moreover, OA is the radial line extending from the center of sight to the point of reflection, so the image of S lies on OA. H [where the two lines intersect] is therefore the image of S, and it is shown in this way that T is the image of I. 
◉ Et continuemus CL. Quia ergo C convertitur ad O ex circumferentia B, erit ymago Q in linea CL. Et OB est linea radialis que extenditur inter visum et punctum conversionis; ergo ymago C est in linea OB. Ergo ymago C est in puncto sectionis inter QL et OB. 
◉ Let us then connect CL. Accordingly, since [the form of] C is reflected to O from the periphery [of the circle passing through] B, [its] image Q will lie on line CL. And OB is the radial line extending between the center of sight and the point of reflection, so the image of C lies on line OB. Consequently, the image of C lies at the intersection of [cathetus] QL and [line of reflection] OB [i.e., at Q]. 
◉ Sed in capitulo de ymagine, cum tractavimus de ymaginibus speculorum spericorum concavorum, patuit quod ymago puncti cuius forma convertitur a concavitate circuli forte concurret cum linea radiali que est inter visum et punctum conversionis ultra circulum, et forte inter visum et circulum, et forte in centro visus, et forte ultra centrum visus, et forte CL equidistans erit OB. 
◉ When we dealt with images in concave spherical mirrors in chapter [2, book 5] on image [formation], however, it was shown [in proposition 32, in Smith, Alhacen on the Principles, 446448] that the image of a point whose form is reflected from the concavity of a [great] circle [on the mirror’s surface] may intersect the radial line linking the center of sight and the point of reflection beyond the circle, or between the center of sight and the circle, or at the center of sight [itself], or behind the center of sight, or CL may be parallel to OB. 
◉ Et in illo capitulo patuit quod forte ymago erit unum punctum, aut duo, aut tria, aut quattuor Ymago ergo C forte erit in BQ, forte ultra Q, et forte in BO, et forte in O, et forte ultra O. Et forte ymago C erit unum punctum, aut duo, aut tria, aut quattuor. 
◉ In that [same] chapter [on image formation], moreover, it was shown that the image might consist of a single point, or of two, or of three, or of four.⁑ So the image of C might lie [at some point between B and Q] on BQ, or perhaps beyond Q, or perhaps on BO, or perhaps at O, or perhaps behind O.⁑ Moreover, the image of C might consist of a single point, or of two, or of three, or [of] four. 
◉ Si ergo ymago C fuerit Q, tunc HT erit dyameter ymaginis SI. Si ergo omnes ymagines SI fuerint in linea HT, tunc forma eius erit linea recta. Sin autem, erit prope rectam, nam medium eius est in rectitudine duarum extremitatum. Si autem ymago C fuerit ultra Q, tunc ymago SI erit fere concava ex parte visus. Et si ymago visus fuerit in linea BO, tunc ymago SI erit convexa ex parte visus. 
◉ Accordingly, if the image of C lies at Q, then HT will be the crosssection of SI’s image. So, if all the images of [points on] SI lie on line HT, its form will be a straight line. If not, however, it will be nearly straight because its midpoint lies on a straight line between two endpoints.⁑ Nevertheless, if the image of C lies beyond Q, the image of SI will be somewhat concave with respect to the center of sight. And if the image of the visible [point C] lies on line BO [i.e., in front of Q], then the image of SI will be convex with respect to the center of sight. 
◉ Et si ymago C fuerit plura puncta, tunc ymago C erit plures linee quarum omnium extremitates coniunguntur in duobus punctis H, T, et media eorum sunt distincta separata. Et HT est dyameter ymaginis SI, quocumque modo fuerit ymago, et dyameter est communis omnibus ymaginibus eius si plures habuerit ymagines, et linea HT est maior quam SI modica quantitate. 
◉ Moreover, if the image of C consists of several points, the image of C will lie on several lines, all of whose endpoints converge at the two points H and T, and their midpoints are distinct and separate. In addition, HT forms the crosssection of image SI, no matter how that image is formed, and [this] crosssection is common to all of its images if it has several images, and line HT [on the image] is longer than [line] SI [on the object] by some amount.⁑ 
◉ Patet ergo quod, cum linee recte equidistantes axi columpnali speculi concavi fuerit in aliquo visibili, ymago eius forte erit recta aut concava, et forte una erit aut plures. 
◉ It is therefore evident that, when straight lines parallel to the axis of a concave cylindrical mirror lie on some visible object, the image of any [one of them] may be straight or concave, and it may consist of a single [line] or [of] several. 
◉ [PROPOSITIO 34] Item iteremus secundam figuram de fallaciis speculorum columpnalium convexorum. In hac autem figura [FIGURE 6.7.34, p. 325] dictum est quod due linee EB, BH convertuntur secundum angulos equales; et quod due linee EG, GT convertuntur secundum angulos equales; et quod HB, TG perveniunt ad L; et HB continet cum BO angulum acutum. Ergo HB secat superficiem contingentem superficiem columpne in B; BL ergo est sub concavitate columpne. Et similiter GL, et similiter due linee BR, GY. 
◉ [PROPOSITION 34] Now let us recapitulate the second diagram [provided in the analysis] of misperceptions in convex cylindrical mirrors [i.e., figure 6.5.19, p. 128, which accompanies proposition 19]. In this proposition [represented by figure 6.8.34, p. 147, abstracted from figure 6.5.19], it has been shown: that the two lines EB and BH are reflected at equal angles; that the two lines EG and GT are reflected at equal angles; that HB and TG converge at L; and [that] HB forms an acute angle with BO. Consequently, HB intersects the plane tangent to the surface of the cylinder at B, so BL lies inside the concavity of the cylinder. And the same holds for GL, as well as for the two lines BR and GY. 
◉ Et duo anguli LBD, DBR sunt equales, et duo anguli LGD, DGY sunt equales. Si ergo RY fuerit in aliquo visibili, et visus fuerit in L, et superficies concava columpne fuerit tersa, tunc forma R extenditur per RB, et pervenit ad B. Et convertetur per BL, et perveniet ad L, et comprehendetur ex L. Et linea HU est perpendicularis super lineam contingentem sectorem ex cuius circumferentia convertentur due linee RB, BL. H ergo est ymago R. Et similiter declarabitur quod forma Y extenditur per YG et convertitur per GL, et ymago eius est T. 
◉ Moreover, the two angles LBD and DBR are equal, and the two angles LGD and DGY are equal. Hence, if RY lies on some visible object, if the center of sight lies at L, and if the concave surface of the cylinder is polished [and therefore reflective], the form of R is extended along RB and reaches B. It will then be reflected along BL and will reach L, and it will be perceived by L. Moreover, line HU [i.e., the cathetus dropped from R through the mirror’s surface] is perpendicular to a line tangent to the [elliptical] section from whose periphery the two lines RB and BL will be reflected. Therefore, H is the image of R. Likewise, it will be proven that the form of Y is extended along YG and is reflected along GL, and its image is T. 
◉ Et continuemus KU. Secabit ergo RY in M. M ergo est in superficie transeunte per axem et per L, nam L et K sunt in hac superficie; ergo KU est in hac superficie. Et quia duo puncta M, L sunt in superficie transeunte per axem columpne, ideo forma M convertetur ad L in hac superficie. Et linea AZ est differentia communis inter superficiem columpne et superficiem transeuntem per suum axem et per L; forma ergo M convertetur ad L per AZ. 
◉ Let us now draw KU [in figure 6.8.34a, p. 148].⁑ Accordingly, it will intersect RY at M. M therefore lies in the plane passing through the axis and through L, for L and K lie in that plane, so KU lies in that plane. Moreover, since the two points M and L lie in a plane passing through the cylinder’s axis, the form of M will be reflected to L within that plane. Line AZ is the common section of the cylinder’s surface and the plane passing through its axis and through L, so the form of M will be reflected to L from [a point on] AZ. 
◉ Et continuemus EM, que est in hac superficie. Et EL etiam est in hac superficie, et punctum E est elevatum a superficie contingente superficiem columpne in linea AZ. Ergo si AZ extrahatur recte in parte Z, concurret cum duabus lineis EM, EL. Concurrat ergo cum EM in I, et cum EL in N. N ergo est inter duo puncta E, L, quia L est intra concavitatem columpne, et N est in superficie columpne, et E est elevatum a columpna. 
◉ Let us then connect EM, which lies in this plane. EL also lies in this plane, and E lies above the plane tangent to the surface of the cylinder along line AZ. Hence, if AZ is extended in a straight line on the side of Z, it will intersect the two lines EM and EL. Accordingly, let it intersect EM at I and EL at N. N therefore lies between the two points E and L because L lies inside the concavity of the cylinder, whereas N lies on the cylinder’s surface, and E lies above the [surface of the] cylinder. 
◉ Et in demonstratione huius figure patuit quod circulus BZG est medius inter lineam HT et superficiem exeuntem ex E equidistantem basi columpne. Et perpendicularis que exit ex E super AZ est in superficie exeunte ex E equidistante basi columpne. Ergo perpendicularis que exit ex E super lineam AZN cadit extra triangulum EIN et in parte N. Angulus ergo EIN est acutus; ergo angulus MIA est acutus. 
◉ Furthermore, in the proof based on this diagram, it was shown that circle BZG lies halfway between [the plane passing through] line HT [parallel to the base of the cylinder] and the plane passing through E parallel to the base of the cylinder. In addition, the perpendicular [EX’] that passes from E through AZ lies in the plane passing through E parallel to the cylinder’s base. Therefore, the perpendicular passing through E to line AZN falls outside triangle EIN and on the side of N. Consequently, angle EIN is acute; so [vertical] angle MIA is [also] acute.⁑ 
◉ Extrahamus ergo ex M perpendicularem super AI, et sit MQ. Q ergo erit ultra I respectu N. Et extrahamus MQ ex parte Q, et dividamus QS ad equalitatem QM. S ergo erit extra superficiem speculi et ultra concavitatem eius, et L erit sub concavitate eius. 
◉ So let us extend MQ [in figure 6.8.34b, p. 149] from M perpendicular to AI. Q will therefore lie beyond I with respect to N [i.e., below I on AZA’]. And let us extend MQ on the side of Q, and let us cut off QS equal to QM. S will therefore lie beyond the surface of the mirror and outside its concavity, while L will lie inside its concavity. 
◉ Et continuemus LS. Secabit ergo NQ in F, et ex F extrahamus FX ad equidistantiam QM. Ergo est perpendicularis super AN et in superficie transeunte per axem et per L; ergo est dyameter circuli exeuntis ex F equidistantis basi columpne. Linea ergo XF est perpendicularis super superficiem contingentem columpnam transeuntem per AZ. 
◉ Let us then draw LS. Accordingly, it will intersect NQ at F, and from F let us extend FX parallel to QM. It is therefore perpendicular to AN and [lies] in the plane passing through the axis and through L, so it is a diameter of the circle [produced on the mirror’s surface by the plane] passing through F parallel to the cylinder’s base. Therefore, line XF is perpendicular to the plane tangent to the cylinder and passing along AZ. 
◉ Et continuemus MF. Erit ergo equalis FS, et duo anguli qui sunt in M, S erunt equales. Et quia XF est equidistans MS, erunt duo anguli F equales duobus angulis qui sunt apud S, M. Due ergo linee MF, FL convertuntur per angulos equales, et XF est perpendicularis super superficiem contingentem superficiem speculi in F. Forma ergo M extenditur per MF, et convertitur per FL, et ymago eius erit S. 
◉ Let us connect MF. Accordingly, it will be equal to FS, and [so] the two angles [FMQ and FSQ] at M and S will be equal [because triangles FMQ and FSQ are equal, by Euclid, I.4]. Moreover, because XF is parallel to MS, the two angles [LFX and MFX] at F will be equal to the two angles [FSQ and FMQ] that are at S and M.⁑ The two lines MF and FL are therefore reflected at equal angles, and XF is perpendicular to the plane tangent to the mirror’s surface at F. So the form of M is extended along MF and is reflected along FL, and its image will be S. 
◉ Et quia due linee RY, HT sunt equidistantes et perpendiculares super superficiem transeuntem per axem et per L (quia HT fuit posita talis), ideo due superficies exeuntes a duabus lineis HT, RY erunt equidistantes. Et quia RY est perpendicularis super superficiem transeuntem per axem et per L, ideo superficies duarum linearum RM, MS erit perpendicularis super superficiem transeuntem per axem et per L. Et erit MS differentia communis hiis duabus superficiebus, et quia AQ est in superficie transeunte per axem, et est perpendicularis super MS, que est differentia communis inter superficiem transeuntem per axem et inter superficiem duarum linearum RM, MS, erit AN perpendicularis super superficiem duarum linearum RM, MS. 
◉ Moreover, since the two lines RY and HT are parallel and [therefore] perpendicular to the plane passing through the axis and through L (because HT was posited as such [in proposition 19]), the two planes passing through the two lines HT and RY [perpendicular to the axis] will be parallel. Since, moreover, RY is perpendicular to the plane passing through the axis and through L, the plane [consisting] of the two lines RM and MS will be perpendicular to the plane passing through the axis and through L. Furthermore, MS will be the common section of these two planes [i.e., RMS and ELDS in figure 6.8.34b], and since AQ lies in the plane [ELDS] passing through the axis and is perpendicular to MS, which is the common section of the plane [ELDS] passing through the axis and the plane [consisting] of the two lines RM and MS, AN will be perpendicular to the plane [consisting] of the two lines RM and MS. 
◉ Et linea AN est equidistans axi columpne; ergo axis columpne est perpendicularis super superficiem in qua sunt due linee RM, MS. Superficies ergo ista est perpendicularis super axem columpne. S ergo in superficie exeunte ex linea RY perpendiculariter super axem columpne. 
◉ But line AN is parallel to the axis of the cylinder, so the axis of the cylinder is perpendicular to the plane containing the two lines RM and MS. Therefore, this plane is perpendicular to the axis of the cylinder. S therefore lies in the plane passing through line RY perpendicular to the axis of the cylinder. 
◉ Sed linea HT est in superficie perpendiculari super axem columpne equidistanti superficiei exeunti ex linea RY. S ergo est extra HT et propinquior L quam HT. Et duo puncta H, T sunt ymagines R, Y, et punctum S est ymago M; ymago ergo linee RMY est linea transiens per H, S, T. 
◉ But line HT lies in a plane perpendicular to the axis of the cylinder and parallel to the plane passing through line RY. Hence, S lies outside [and above] HT and nearer to L than HT. In addition, the two points H and T are the images of R and Y, and point S is the image of M, so the image of line RMY is the line passing through H, S, and T. 
◉ Sed talis linea est arcualis, quia S est extra HT, et transeat per puncta H, S, T linea HST arcualis. Et quia HT, secundum positionem, fuit elevata a convexo columpne, erit HT ultra superficiem speculi respectu L. Et iam declaravimus quod S est ultra concavitatem speculi respectu L; ergo tota linea HST est ultra concavitatem superficiei speculi. Et L est sub concavitate speculi; ergo L est extra superficiem in qua est linea HST. Arcualitas ergo linee HST apparebit visui L manifeste. 
◉ Such a line is curved, however, because S lies outside HT, and a curved line HST must pass through points H, S, and T. And since HT lay beyond the convex [side of the] cylinder, according to construction, HT will lie beyond the surface of the mirror with respect to L. In addition, we have already shown that S lies beyond the concavity of the mirror with respect to L, so the entire line HST lies beyond the concavity of the mirror’s surface. Moreover, L lies inside the concavity of the mirror, so L lies outside the plane containing line HST. Therefore, the curvature of line HST will appear clearly to the eye at L. 
◉ Et quia F est in superficie columpne, et TH est ultra columpnam, et TH est in superficie trianguli LHT, erit linea LFS altior quam superficies trianguli LHT. Linea ergo LS erit altior duabus lineis LH, HT respectu visus L. S ergo est altior quam dua puncta H, T; linea ergo HST apparebit visui L concava. 
◉ Furthermore, since F lies on the surface of the cylinder, while TH lies beyond the cylinder, and since TH lies in the plane of triangle LHT, line LFS will be higher than the plane of triangle LHT. Line LS will therefore be higher than the two lines LH and HT with respect to the center of sight at L. Consequently, S is higher than the two points H and T, so line HST will appear concave to the center of sight at L. 
◉ [PROPOSITIO 35] Item secemus columpnam per superficiem declinem super axem eius, et non transeat per totum axem. Faciet ergo sectorem. Sit ergo ABG [FIGURE 6.8.35, p. 326]. Sed in prima figurarum de columpnis concavis declaratum est quod in superficie cuiuslibet sectoris columpne erit perpendicularis super superficiem contingentem columpnam ex cuius extremitatibus convertuntur forme. Sit ergo perpendicularis GZ, et sit BE perpendicularis super lineam contingentem circumferentiam sectoris in B, et sit B prope G. BK ergo secabit perpendicularem GZ, et continebit cum ipsa angulum acutum. Secet ergo in E. Angulus ergo BEG erit acutus. 
◉ [PROPOSITION 35] To continue, let us cut the cylinder with a plane slanted to its axis, but do not let it pass through the entire axis [so as to cut the cylinder along a line of longitude]. Accordingly, it will form an [elliptical] section. Let it therefore be ABG [in figure 6.8.35, p. 150].⁑ But in the first of the propositions concerning concave cylindrical [mirrors] it was demonstrated that in the plane of any [elliptical] section on a cylinder there will be a normal to the plane tangent to the cylinder from whose endpoints forms are reflected.⁑ So let that normal be GZ[A], let BE[K] be perpendicular to the line tangent to the periphery of the [elliptical] section at B, and let B lie near G. BK will therefore intersect normal GZ, and it will form an acute angle with it. Accordingly, let it intersect at E. Angle BEG will therefore be acute [by proposition 16, lemma 5]. 
◉ Et extrahamus ex G lineam ad equidistantiam linee BK, et sit GD. Angulus ergo DGE erit acutus; ergo GD erit intra concavitatem columpne. Et ponamus angulum EGL equalem angulo EGD. GL ergo concurret cum BE in L. Et signemus M in linea LE. Erit ergo angulus MAG acutus, quia AM est intra sectorem. 
◉ From G let us extend line GD parallel to line BK. Hence, angle DGE will be acute, so GD will lie inside the concavity of the cylinder. Let us then take angle EGL equal to angle EGD. GL will thus intersect BE at L. And let us mark M on line LE [inside the elliptical section]. Angle MAG will therefore be acute because AM lies inside the [elliptical] section. 
◉ Et ponamus angulum GAD equalem angulo GAM. Ergo AD concurret cum GD, nam duo anguli qui sunt apud A, G sunt acuti. Concurrant ergo in D. AD ergo secabit BK. Secet ergo in T. 
◉ Let us then take angle GAD equal to angle GAM. Therefore, AD will intersect GD, since the two angles [GAD and AGD] that are at A and G are acute. So let them intersect at D. AD will therefore intersect BK. Let it then intersect at T. 
◉ Cum ergo BK fuerit in aliquo visibili, et visus fuerit in D, tunc forma L videbitur in G, quia forma L convertetur ad D ex G, et quia DG est equidistans perpendiculari LB. Et forma M videtur in T, quia forma M convertitur ad G ex A, et T est ymago M. 
◉ Consequently, if BK lies on some visible object, and if the center of sight lies at D, the form of L will be seen at G because the form of L will be reflected to D from G, and DG is parallel to normal LB [by construction].⁑ Meantime, the form of M is seen at T because the form of M is reflected to G from A, and T is the image of M. 
◉ Et transeat per D superficies equidistans basi columpne. Secabit ergo superficiem ABG et faciet in superficie columpne circulum COR. Superficies ergo huius circuli secabit BK, secat enim GD, que est ei equidistans. Secet ergo BK in K, et sit centrum circuli CR punctum H. Et continuemus DH, et transeat ad R. Et continuemus KH, et transeat ad C. 
◉ Now let a plane pass through D parallel to the base of the cylinder [to form the circle represented at the bottom of figure 6.8.35]. Accordingly, it will intersect the plane of ABG and will form circle COR on the surface of the cylinder. The plane of this circle will therefore intersect BK, for it intersects GD, which is parallel to it [by construction]. Let it therefore intersect BK at K, and let point H be the center of circle CR. Let us then draw DH, and let it pass to R. Let us also draw KH, and let it pass to C. 
◉ Forma ergo K convertitur ad D ex circumferentia ex arcu RC, ut patuit in ymaginibus circulorum. Convertatur ergo ex O, et continuemus KO, DO, HO. Anguli ergo qui sunt apud O sunt equales, et DO secabit HC in N. N ergo est ymago K. 
◉ Hence, the form of K is reflected to D from the periphery [of the circled centered on H] within arc RC, as was shown in [the analysis of] images [formed] in [great] circles [within concave spherical mirrors].⁑ So let it be reflected from O, and let us draw KO, DO, and HO. The angles [DOH and KOH] at O are therefore equal, and [reflected ray] DO will intersect [cathetus K]HC at N. So N is the image of K. 
◉ Et continuemus KD. KD ergo erit differentia communis inter circulum RC et sectorem ABG, nam duo puncta K, D sunt in utraque superficie, nichil enim de superficie sectoris ABG est in superficie circuli RC nisi linea KD. G ergo est extra circulum, et similiter T, et sunt in superficie sectoris. 
◉ Let us then connect KD. Accordingly, KD will be the common section of circle RC and [elliptical] section ABG, since the two points K and D lie in both planes, for there is nothing except line KD in plane ABG of the [elliptical] section that is [also] in the plane of circle RC. G therefore lies outside the circle, and likewise T, and both lie in the plane of the [elliptical] section. 
◉ Et N est in superficie circuli, et forma LMK transit per puncta G, T, N, et linea que transit per hec puncta est arcualis. Sed superficies sectoris est declinis super superficiem columpne; axis ergo sectoris non transit per totam axem columpne, nec est equidistans basi columpne. 
◉ N, meanwhile, lies in the plane of the circle, and the form of LMK passes through points G, T, and N, and [so] the line that passes through these points is curved. However, the plane of the [elliptical] section is slanted with respect to the cylinder’s surface, so the [major] axis of the [elliptical] section does not pass along the entire axis of the cylinder, nor is it parallel to the cylinder’s base. 
◉ Patet ergo ex hac figura et duabus premissis quod linee recte equidistantes axi columpne et equidistantes basi eius, et etiam ille que declinantur super superficiem eius, forte videbuntur arcuales, forte recte, forte converse. Item, quia T est ymago M, et N ymago K, erit forma MK conversa. 
◉ From this and the previous two propositions it is therefore evident that straight lines parallel to the axis of the cylinder, as well as those parallel to its base, and also those that are slanted with respect to its surface may appear curved, or straight, or reversed. Furthermore, since T is the image of M and N the image of K, the form of MK will be reversed. 
◉ Et si linea etiam fuerit in superficie circuli equidistanti basi columpne, cuius superficies transit per centrum visus, ut dictum est in ymaginibus circulorum in septimo capitulo huius tractatus, forma forte erit equalis recta, forte conversa. 
◉ In addition, if the line also lies in the plane of a circle parallel to the cylinder’s base, and if the plane of that circle passes through the center of sight, the image may be the same size [as its object] and properly oriented, or it may be reversed, as was claimed in [propositions 2527 of] the seventh chapter of this book [dealing] with images in [great] circles [on concave spherical mirrors]. 
◉ Patet ergo quod forma eorum que comprehenduntur in speculis columpnalibus concavis forte erit recta, forte conversa. 
◉ It is thus evident that the forms of objects perceived in concave cylindrical mirrors may be properly oriented, or they may be reversed. 
◉ [PROPOSITIO 36] Item iteremus formam tertie figure de fallaciis speculorum spericorum concavorum, ipsis litteris existentibus. Et sit circulus BZA [FIGURE 6.8.36, p. 327] in superficie speculi columpnalis concavi, et sit visus in D. Erit ergo extra superficiem circuli, et erunt due linee EA, EB perpendiculares super superficiem contingentem superficiem columpne. Et erit superficies trianguli DGE perpendicularis super superficiem circuli, quia DG est perpendicularis super superficiem circuli. 
◉ [PROPOSITION 36] Now let us recapitulate the diagram for the third proposition [dealing] with misperceptions in concave spherical mirrors, leaving the letters as they are [in figure 6.8.36, p. 151, which combines figures 6.7.26 and 6.7.27]. Let BZA be a circle on the surface of a concave cylindrical mirror, and let the center of sight be at D [on DG, which is perpendicular to the plane of BZA]. It will therefore lie outside the circle’s plane, and the two lines EA and EB will [each] be perpendicular to a plane tangent to the cylinder’s surface [at points A and B]. In addition, the plane of triangle DGE will be perpendicular to the plane of the circle because DG is perpendicular to the plane of the circle. 
◉ Superficies ergo trianguli DGE transit per totum axem et per D, et neutra superficies DBO, DAO, que se secant in linea DO, transit per totum axem. Et in neutra superficie est aliquid de axe columpne nisi E, quod est centrum circuli. Et utraque superficies DBO, DAO facit in superficie columpne sectorem, et forme convertuntur ex hiis sectoribus a duobus punctis A, B. 
◉ Hence, the plane of triangle DGE passes through the entire axis as well as through D, whereas neither plane DBO nor plane DAO, which intersect along line DO, passes through the entire axis. Moreover, there is nothing but E on the cylinder’s axis in either plane, E being the circle’s center. And each of the planes DBO and DAO forms an [elliptical] section on the cylinder’s surface, and forms are reflected from these [elliptical] sections at the two points A and B. 
◉ Forma ergo R convertitur ad D ex B, et forma M convertitur ad D ex A, et NU erit dyameter ymaginis MR, et est minor quam MR. Et similiter duo puncta H, L convertuntur ad D ex duobus punctis A, B, et erit TK dyameter ymaginis LH, et est ei equalis. Et erit CI diameter ymaginis FQ, et est maior illa. Et omnes iste ymagines erunt converse. 
◉ The form of R is therefore reflected to D from B, whereas the form of M is reflected to D from A, and NU will be the crosssection of the image of MR, and it is shorter than MR. Likewise, the [forms of the] two points H and L are reflected to D from the two points A and B, and TK will be the crosssection of LH’s image, and it is the same size as TK. Finally, CI will be the crosssection of FQ’s image, and it is longer than FQ. All of these images, moreover, will be reversed. 
◉ Et si visus fuerit in O, et linee CI, TK, NU fuerint visibiles, erunt econtra, tunc enim dyameter ymaginis CI erit minor ipsa, et dyameter ymaginis NU erit maior ipsa, et erit dyameter TK equalis ei, et omnes ymagines erunt recte. Et omnia ista ostensa sunt in predicto capitulo. 
◉ But if the center of sight lies at O, and if lines CI, TK, and NU are the visible objects, the opposite will obtain, for in that case the crosssection of the image [FQ] of CI will be shorter than CI, whereas the crosssection of the image [MR] of NU will be longer than NU, and the crosssection TK [of LH’s image] will be the same size as it, and these images will all be properly oriented. All these points were shown in the preceding chapter. 
◉ Item cum utraque extremitas alicuius harum habuerit unam ymaginem, et aliquod punctum in medio habuerit plures ymagines, tunc illa linea habebit tot ymagines quot punctum medium habet. Et si utraque extremitas vel altera habuerit plures ymagines, et punctum medium habuerit unam, tunc linea tot habebit ymagines quot habuerit punctum extremum. Et si utraque extremitas aut altera habuerit multas ymagines, et punctum medium habuerit multas ymagines, tunc linea habebit ymagines secundum maiorem numerum. Et hoc patebit ut de ymaginibus patuit speculorum spericorum concavorum. 
◉ Furthermore, when either endpoint of any of these [lines] has a single image, and when any intermediate point [on that line] has several images, that line will have as many images as the intermediate point has. If, moreover, one endpoint or the other [of the line] has several images, and if the intermediate point has one, then the line will have as many images as the endpoint has. And if one endpoint or the other has several images, and if the intermediate point has several images, the line will yield images according to the greatest number [as pointed out in note 130, p. 256]. This will be shown as was shown for images in concave spherical mirrors. 
◉ In speculis ergo columpnalibus concavis accidit fallacia in omnibus que in eis comprehenduntur sicut accidit in speculis spericis concavis, scilicet de formis specierum visibilium, et de quantitatibus et de numero suarum ymaginum, et de rectitudine et de conversione, cum fallaciis que appropriantur conversioni. Et fallacie erunt in hiis ut in speculis predictis, et hec sunt que voluimus declarare in capitulo hoc. 
◉ Hence, in concave cylindrical mirrors misperception occurs in all respects as it occurs in concave spherical mirrors, that is, concerning the shapes of visible forms, concerning the sizes and number of their images, and concerning their proper orientation or reversal, along with the misperceptions that apply to reflection [itself]. And the misperceptions in these cases will be as they are in the aforementioned mirrors, and these are the points we wanted to demonstrate in this chapter. 
◉Capitulum nonum 
◉Chapter Nine 
De fallaciis que accidunt in speculis piramidalibus concavis 

◉ In hiis autem accidunt ille fallacie que accidunt in speculis columpnalibus concavis. Debilitas vero coloris et lucis, et diversitas positionis et remotionis accidunt in hiis sicut in omnibus speculis, nam causa huius est conversio. Accidit etiam in hiis speculis multitudo ymaginum, sicut in speculis columpnalibus et spericis concavis, sicut dictum est in capitulo de ymaginibus. Accidit etiam in eis ut columpnalibus concavis, scilicet quod rectum videtur convexum et videtur concavum. 
◉ In these [sorts of mirrors] the misperceptions that occur are those that occur in concave cylindrical mirrors. Indeed, the weakening of color and light, as well as variation in location and distance, occur in these mirrors as in all [other kinds of] mirrors, for the cause of this is reflection [itself]. In addition, a multitude of images arises in these mirrors, just as in concave cylindrical and spherical mirrors, as was claimed in chapter [2, book 5] on image [formation]. What happens in these mirrors is also like what happens in concave cylindrical [mirrors], i.e., what is straight appears convex, or it appears concave. 
◉ Huius autem demonstratio est quia linee recte que extenduntur in longitudine speculi que transit per caput piramidis, et que sunt prope illas, videntur convexe, et videntur concave, et forte recte. 
◉ The demonstration of this is that straight lines that extend along the length of the mirror and pass through the cone’s vertex, as well as those that are near these [lines in orientation], appear convex, or they appear concave, or perhaps [they appear] straight. 
◉ [PROPOSITIO 37] Et demonstratio super hoc est ut demonstratio in speculis columpnalibus concavis, nam si iteraverimus secundam figuram de fallaciis speculorum piramidalium convexorum, inveniemus dyametrum ymaginis linee recte posite in illo speculo, qui est illic linea AY intra concavitatem speculi piramidalis, et inveniemus punctum quod est sub superficie contingente piramidem transeuntem per lineam ex qua convertitur forma linee recte ad visum, quod illic punctum F. 
◉ [PROPOSITION 37] The demonstration of this point is like the demonstration [given] for concave cylindrical mirrors [in proposition 33, pp. 221224 above], for if we recapitulate the second diagram concerning misperceptions in convex conical mirrors [i.e., the top diagram of figure 6.6.22a, p. 131, from which figure 6.9.37, p. 152, has been abstracted], we will find the crosssection of the image of straight line [AN] placed toward that mirror, which, in that case, is [curved line] A[P]Y inside the concavity of the conical mirror, and we will find the point that is below the plane tangent to the cone and passing along the line [AZE] from which the form of the straight line [AN] is reflected to the center of sight, which is F in that case. 
◉ Si fuerit punctum centrum visus, erunt omnia puncta que sunt in dyametro ymaginis conversa ad punctum F, et ymagines duarum extremitatum A, Y erunt extremitates linee recte AN, et loca ymaginis puncti quod est in medio AY diversabuntur. Et hoc declarabitur eadem via qua processimus in demonstratione prime figure speculorum columpnalium concavorum. 
◉ If [that] point is the center of sight, all the points that are on the crosssection of the image will be reflected to point F, and the images of the two endpoints A and Y [on objectline APY] will be the endpoints of straight line AN, and the imagelocation of intermediate [point P] on AY will vary. And this will be demonstrated by the same train [of logic] we followed in the proof [provided] in the first proposition [dealing] with concave cylindrical mirrors [i.e., proposition 33].⁑ 
◉ Patet ergo ex hoc quod si AY fuerit in aliquo visibili, et visus fuerit F, tunc ymago forte videbitur convexa, et forte concava. Et patet etiam in secunda figura de fallaciis speculorum columpnalium concavorum quod linee posite in latitudine speculi apparebunt concave concavitate mirabili, et quod ymagines linearum rectarum que sunt in superficiebus transeuntibus per axem et per centrum visus erunt recte. 
◉ From this it is clear that, if AY lies on some visible object, and if F is the center of sight, the image may appear convex, or it may appear concave.⁑ And it is also evident from the second proposition concerning misperceptions in concave cylindrical mirrors [i.e., proposition 34] that lines placed along the width of a [concave conical] mirror will appear concave with a remarkable curvature and that images of straight lines that lie in planes passing through the axis and the center of sight will be straight. 
◉ [PROPOSITIO 38] Item iteremus tertiam figuram de fallaciis speculorum spericorum concavorum eisdem litteris. Si ergo aliquod punctum fuerit in axe piramidis, et due linee EA, EB fuerint perpendiculares super superficies contingentes piramidem (et hoc est possibile, quia sunt equales, possunt enim cum axe continere duos angulos acutos equales), cum ergo hee due linee fuerint perpendiculares, et visus fuerit D, tunc superficies in qua sunt linee GE, ED transibit per totum axem et per centrum visus. 
◉ [PROPOSITION 38] Now with the same letters, let us recapitulate the third figure concerning misperceptions in concave spherical mirrors [as given in figure 6.8.36, p. 151, which combines figures 6.7.26 and 6.7.27]. If, therefore, some point [i.e., E] lies on the axis of the cone, and if the two lines EA and EB [passing through that point] are perpendicular to planes tangent to the cone (and this is possible because they are equal, since they can form two equal acute angles with the axis), then when [each of] these two lines is perpendicular [to a plane tangent to the mirror], and when the center of sight is at D, the plane containing lines GE and ED will pass through the entire axis as well as through the center of sight. 
◉ Et utraque superficies DAM, DBR erit declinis super axem piramidis, et erunt differentie earum due sectores piramidis. Et erit forma punctorum R, H, Q conversa ad D ex B, et forme punctorum L, M, F convertuntur ad D ex A. Cum ergo linee MR, LH, FQ fuerint in aliqua superficie visibili, et visus fuerit in D, tunc NU erit ymago MR, et TK erit ymago LH, et CI erit ymago FQ. 
◉ Furthermore, both planes [containing] DAM and DBR will be inclined to the axis of the cone, and [so] the common sections of those two [planes and the mirror’s surface] will be conic sections. The form[s] of points R, H, and Q will be reflected to D from B, and the forms of points L, M, and F are reflected to D from A. Hence, if lines MR, LH, and FQ lie on some visible surface, and if the eye is at D, then NU will be the image of MR, TK will be the image of LH, and CI will be the image of FQ. 
◉ Sic ergo ymago MR erit minor se ipsa, et ymago FQ maior se ipsa, et ymago LH equalis sibi ipsi, et omnes ymagines erunt converse. 
◉ Thus, the image [NU] of MR will be shorter than [MR] itself, the image [CI] of FQ will be longer than [FQ] itself, and the image [TK] of LH will be the same size as [LH] itself, and all the images will be reversed. 
◉ Et si visus fuerit in O et NU, TK, CI fuerint in superficiebus visibilium, tunc ymagines earum erunt MR, LH, FQ. Sic ergo erit ymago CI minor se ipsa, et ymago NU maior, et ymago TK equalis. 
◉ If, moreover, the center of sight is at O, and NU, TK, and CI are on the surfaces of visible objects, their images will be MR, LH, and FQ. Accordingly, the image [FQ] of CI will be shorter than [CI] itself, the image [MR] of NU [will be] longer [than NU itself], and the image [LH] of TK will be the same size [as TK itself]. 
◉ Et iste ymagines erunt recte, nam iste ymagines erunt ultra centrum visus et comprehendentur ante visum super lineas radiales. Puncta ergo M, L, F comprehenduntur in linea AO, et puncta R, H, Q comprehenduntur in OB, et sic forma revertetur recta. 
◉ And these images will be properly oriented, for these images will lie behind the center of sight and will be perceived facing the center of sight along [direct] radial lines.⁑ Consequently, points M, L, and F are perceived along line [of reflection] AO, whereas points R, H, and Q are perceived along [line of reflection] OB, and so their form will be reflected with proper orientation. 
◉ Patet ergo ex hiis que diximus in hoc capitulo quod linee recte quandoque videntur in hiis speculis convexe, quandoque concave, quandoque recte, et quandoque maiores, et minores, et equales, et quandoque recte, et converse. 
◉ From what we have claimed in this chapter, therefore, it is clear that straight lines sometimes appear convex in these mirrors, sometimes concave, and sometimes straight, and [they] sometimes [appear] longer, [sometimes] shorter, and [sometimes] the same size [as they actually are], and [they sometimes appear] properly oriented, and [sometimes] reversed. 
◉ Et in capitulo de ymagine declaravimus quod omne punctum visibile in huiusmodi speculis quandoque habet unam ymaginem, quandoque duas, et tres, et quattuor. Omnia ergo que comprehenduntur in huiusmodi speculis accidit in eis fallacia ut in columpnalibus concavis, et accidunt in eis etiam fallacie composite sicut in ceteris speculis. Et exempla et declaratio eorum sunt sicut in speculis planis. Et hoc intendimus declarare in hoc capitulo. Nunc autem finiamus sextum tractatum. 
◉ In chapter [2, book 5] on image [formation], moreover, we showed that in mirrors of this kind every visible point sometimes has one image, sometimes two, or three, or four. Therefore, misperception occurs in everything that is perceived in this sort of mirror, just as in concave cylindrical [mirrors], and compound misperceptions also occur in these as in the rest of the mirrors. Examples and proof of these [kinds of compound misperceptions] are as [they can be found] in plane mirrors. And we intended to explain this in this chapter. Now, however, let us end the sixth book. 