§13## ιγ΄.
## 13.
Δυνατόν ἐστι διὰ πλειόνων ἐνόπτρων ἐπιπέδων ἰδεῖν τὸ αὐτό.
It is possible to see the same object through multiple plane mirrors.
ἔστω, ὃ δεῖ ὀφθῆναι, τὸ Α, ὄμμα δὲ τὸ Β, ἔνοπτρα δὲ τρία τὰ Γ∠, ∠Ε, ΕΖ. ἤχθω δὴ κάθετος ἀπὸ τοῦ Β ἐπὶ τὸ Γ∠ ἔνοπτρον ἡ ΒΓ, ἴση δὲ ἡ ΒΓ τῇ ΓΣ. καὶ πάλιν ἀπὸ τοῦ Α ἐπὶ τὸ ΕΖ κάθετος ἡ ΑΖ, καὶ τῇ ΑΖ ἴση ἡ ΖΘ, καὶ ἀπὸ τοῦ Θ ἐπὶ τὸ ∠Ε ἔνοπτρον κάθετος ἤχθω ἡ ΘΚ, καὶ ἔστω τῇ ΘΚ ἴση ἡ ΚΛ, καὶ ἀπὸ τοῦ Λ ἐπὶ τὸ Σ ἐπεζεύχθω ἡ ΛΜΞΣ, ἀπὸ δὲ τοῦ Μ ἐπὶ τὸ Θ ἡ ΜΡΘ, ἐπεζεύχθωσαν δὲ καὶ αἱ ΑΡ, ΒΞ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΒΓ τῇ ΓΣ, καὶ ὀρθαὶ αἱ πρὸς τῷ Γ γωνίαι, δύο δὴ αἰ ΒΓ, ΓΦ δυσὶ ταῖς ΣΓ, ΓΦ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ γωνία ἡ ὑπὸ ΒΓΦ ὀρθὴ οὖσα γωνίᾳ τῇ ὑπὸ Σῶ ὀρθῇ οὔσῃ ἴση ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφʼ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν, ἡ μὲν πρὸς τῷ Β γωνία τῇ πρὸς τῷ Σ, ἡ δὲ γωνία τῇ Τ. ἀλλʼ ἡ Τ τῇ Ν ἐστιν ἴση· κατὰ κορυφὴν γάρ·
Let the object to be seen be A, the eye B, and the three mirrors GD, DE, EZ. Let then the perpendicular BG be drawn from B to the mirror GD, and let BG be equal to GS. And again, let the perpendicular AZ be drawn from A to the mirror EZ, and let AZ be equal to ZTh. And let the perpendicular ThK be drawn from Th to the mirror DE, and let ThK be equal to KL. And let LMXS be joined from L to S. And MRTh from M to Th. And let AR, BX also be joined. Since then BG is equal to GS, and the angles at G are right, the two lines BG, GF are equal to the two SG, GF, each to each. And the angle BGF, being right, is equal to the angle S..., which is also right. And the remaining angles will be equal to the remaining angles, which the equal sides subtend. That is, the angle at B to the angle at S, and the angle [GFB] to T. But T is equal to N (for they are vertical); so that the angle N is also equal to X.
ὥστε ἴση ἐστὶ καὶ ἡ Ν γωνία τῇ Ξ. ἡ ἄρα ΒΞ ὄψις ἀνακλασθήσεται ἐπὶ τὸ Μ. πάλιν ἐπεὶ ἴση ἐστὶν ἡ ΘΚ τῇ Κ Λ, καὶ ὀρθαὶ δὲ αἱ πρὸς τῷ Κ, ἴση ἐστὶν ἡ Ο γωνία τῇ Π. ἀνακλᾶται ἄρα ἡ.
Therefore, the visual ray BX will be reflected to M. Again, since ThK is equal to KL, and the angles at K are also right, the angle O is equal to P.
αὐτὴ ὄψις ἡ ΒΞΜ ἐπὶ τὸ Ρ διὰ τὰ αὐτὰ δὴ καὶ ἐπὶ τὸ Α διὰ τὸ ἴσην εἶναι τὴν ὑπὸ ΖΡΑ γωνίαν τῇ ὑπὸ ΕΡΜ ὁμοίως ταῖς λοιπαῖς ἀποδείξεσιν.
Therefore, the same visual ray BXM is reflected to R, and indeed for the same reasons also to A, because the angle ZRA is equal to the angle ERM, similarly to the other proofs.
ὁρᾷ ἄρα ἡ ἀπὸ τοῦ Β ὄμματος ὄψις τὸ Α διὰ τῶν τριῶν ἐνόπτρων ὄντων ἐπιπέδων τῶν Γ∠, ∠Ε, ΕΖ.
Therefore, the visual ray from the eye B sees A through the three plane mirrors GD, DE, EZ.
§14## ιδ΄.
## 14.
Ἔστι δὲ καί, διʼ ὅσων ἄν τις ἐπιτάξῃ ἐνόπτρων ἐπιπέδων, ἰδεῖν τὸ αὐτό·
It is also possible to see the same object through as many plane mirrors as one may prescribe.
δεῖ δὲ κατὰ τὸν ἀριθμὸν τῶν ἐνόπτρων πολύγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον συνίστασθαι δυσὶ πλείους ἔχον πλευρὰς τῶν ἐνόπτρων.
But it is necessary that, according to the number of mirrors, an equilateral and equiangular polygon be constructed, having two more sides than the mirrors.
ἔστω γάρ, ὃ μὲν ὀφθῆναι δεῖ, τὸ Α, ὄμμα δὲ τὸ Β, καὶ ἐπεζεύχθω ἡ ΑΒ, καὶ ἀπὸ τῆς ΑΒ ἀναγεγράφθω πολύγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον δύο πλευρὰς, πλείους ἔχον τῶν ἐνόπτρων καὶ ἔστω τὸ ΑΒ∠ πολυγώνιον, καὶ εἰλήφθω τὸ κέντρον τοῦ κύκλου τοῦ γραφομένου περὶ τὸ πολύγωνον τὸ Θ, καὶ ἀπʼ αὐτοῦ ἐπεζεύχθωσαν αἱ ΘΓ, ΘΕ, Θ∠, ΘΒ, ΘΑ ἐπὶ τὰς γωνίας, καὶ προσκείσθωσαν ἔνοπτρα ἐπίπεδα πρὸς ὀρθὰς ταῖς ἐπεζευγμέναις.
For let the object to be seen be A, the eye B, and let AB be joined. And on AB let there be described an equilateral and equiangular polygon, having two more sides than the mirrors, and let it be the polygon ABD... And let the center of the circle described about the polygon be Th. And from it let ThG, ThE, ThD, ThB, ThA be joined to the angles. And let plane mirrors be placed at right angles to the joined lines.
ἐπεὶ οὖν ἴση ἐστὶν ἡ Ζ Λ γωνία τῇ ΝΚ ὀρθὴ γάρ ἐστιν ἑκατέρα· ὧν ἡ Ν τῇ Λ ἴση ἐστίν, λοιπὴ ἄρα ἡ Ζ τῇ Κ ἴση ἐστίν.
Since then the angle ZL is equal to NK, for each is a right angle; of which N is equal to L, therefore the remaining Z is equal to K.
ὥστε ἡ ἀνάκλασις τῆς ΒΓ ὄψεως ἐπὶ τὸ ∠ ἔσται· διὰ γὰρ ἴσων γωνιῶν αἱ ἀνακλάσεις γίνονται.
So that the reflection of the visual ray BG will be to D; for reflections occur through equal angles.
ὁμοίως δὲ δειχθήσονται καὶ αἱ πρὸς τοῖς ∠, Ε σημείοις γωνίαι ἴσαι αἱ πρὸς τοῖς ἐνόπτροις.
And similarly the angles at the points D, E with respect to the mirrors will be shown to be equal.
ἡ ἄρα ἀπὸ τοῦ Β ὄμματος ὄψις ἀνακλωμένη καὶ προσπεσοῦσα πρὸς πάντα τὰ ἔνοπτρα ἥξει ἐπὶ τὸ Α.
Therefore, the visual ray from the eye B, being reflected and falling on all the mirrors, will arrive at A.