§28## κη΄.
## 28.
Ἐν τοῖς κοίλοις ἐνόπτροις ἐὰν τὴν ἐκ τοῦ κέντρου δίχα τεμὼν καὶ πρὸς ὀρθὰς ἀγαγὼν θῇς τὰ ὄμματα ἴσον ἀπέχοντα τῆς ἐκ τοῦ κέντρου, θῇς δὲ ἢ ἀνὰ μέσον τῆς διαμέτρου καὶ τῆς πρὸς ὀρθὰς ἢ ἐπʼ αὐτῆς τῆς πρὸς ὀρθάς, οὐδέτερον τῶν ὀμμάτων φαίνεται.
In concave mirrors, if, having bisected the line from the center and erected a perpendicular, you place the eyes at equal distances from the line from the center, and you place them either between the diameter and the perpendicular, or on the perpendicular itself, neither of the eyes appears.
ἔστω κοῖλον ἔνοπτρον τὸ ΑΓ∠, διάμετρος δὲ ἡ Α∠, κέντρον δὲ τὸ Κ, καὶ ἡ πρὸς ὀρθὰς ἡ ΚΓ δίχα τετμήσθω κατὰ τὸ Π, πρὸς ὀρθὰς δὲ αὐτῇ ἔστω ἡ ΕΠΖ, καὶ ὄμματα τὰ Β, Θ μεταξὺ κείμενα τῆς τε διαμέτρου τῆς Α∠ καὶ τῆς ΕΖ ἐν παραλλήλοις ταῖς ΕΖ, ΒΘ ἴσον ἀπέχοντα τῆς ΚΓ, ὄψις δὲ ἔστω ἡ ΒΓ ἀνακλωμένη ἐπὶ τὸ Θ ἴσας γὰρ ποιεῖ γωνίας πρὸς τῇ περιφερείᾳ διὰ τὸ παράλληλον εἶναι τὴν ΖΕ τῇ ΒΘ καὶ ἴσην τὴν ΒΝ τῇ ΝΘ. καὶ ἐπιζευχθεῖσαι αἱ ΚΒ, ΚΘ ἐκβεβλήσθωσαν, ἐκβεβλήσθω δὲ καὶ ἡ ΓΒ ἐπὶ τὸ Φ. καὶ ἐπεὶ μείζων ἐστὶν ἡ ΒΓ τῆς Β Κ, μείζων ἐστὶν ἡ Ρ γωνία τῆς Ι. ὥστε καὶ ἡ ὑπὸ ΓΒΘ μείζων τῆς ὑπὸ ΘΒΚ, τουτέστι τῆς ὑπὸ ΒΘΚ. οὐκ ἄρα συμπεσεῖται ἡ ΒΓ τῇ ΚΘ. οὐκ ἄρα ὀφθήσεται τὸ Θ κατὰ γὰρ τὴν συμβολὴν φαίνεται τῶν ΒΓ, ΚΘ.
ἔστω πάλιν τὰ αὐτὰ τῇ ἐπάνω, τὰ δὲ Β, Θ ὄμματα ἔστωσαν ἐπὶ τῆς δίχα καὶ πρὸς ὀρθὰς τεμνούσης τὴν ἐκ τοῦ κέντρου ἐπὶ τῆς Α∠.
Let the concave mirror be AGD, the diameter AD, the center K, and let the perpendicular KG be bisected at P, and let EPZ be perpendicular to it, and let the eyes B, Th be placed between the diameter AD and EZ, in parallels EZ, BTh, at equal distances from KG, and let the visual ray be BG, being reflected to Th; for it makes equal angles at the circumference because ZE is parallel to BTh and BN is equal to NTh. And let KB, KTh, being joined, be extended, and let GB also be extended to Ph. And since BG is greater than BK, angle R is greater than I. So that angle GBTh is also greater than angle ThBK, that is, than angle BThK. Therefore, BG will not meet KTh. Therefore, Th will not be seen; for it appears at the intersection of BG, KTh. Let the same things as above hold again, and let the eyes B, Th be on the line bisecting and perpendicular to the line from the center on AD.
ἐπεὶ οὖν ἴση ἡ μὲν ΒΓ τῇ ΒΖ, ἡ δὲ ΓΘ τῇ Ζ Θ, παράλληλος ἂν εἴη ἡ ΒΓ τῇ ΖΘ. οὐκ ἄρα συμπεσεῖται ἡ ΒΓ ὄψις τῇ ἐκ τοῦ κέντρου ἐπὶ τὸ ὁρώμενον, τουτέστι τῇ ΖΘ, ἐπὶ τὰ Θ, μέρη.
Since, therefore, BG is equal to BZ, and GTh is equal to ZTh, BG would be parallel to ZTh. Therefore, the visual ray BG will not meet the line from the center to the object, that is, ZTh, on the parts Th.
ὥστε οὐ φαίνεται τὸ Θ ὄμμα· κατὰ γὰρ τὴν συμβολὴν ἐφαίνετο τῶν ΒΓ, ΖΘ.
ἔστω πάλιν τὰ αὐτά, τῆς δὲ διχοτομίας ἀνωτέρω κείσθω τὰ ὄμματα τὰ Β, Γ ἴσον ἀπέχοντα τῆς ἐκ τοῦ κέντρου τῆς Ζ Α. φημὶ δὴ φαίνεσθαι τὰ Β, Γ καὶ τὰ δεξιὰ ἀριστερὰ καὶ τὰ ἀριστερὰ δεξιὰ καὶ τὸ εἴδωλον μεῖζον τοῦ προσώπου καὶ τὸ ἀπόστημα ἀπὸ τοῦ ἐνόπτρου ἔχον μεῖζον τὸ εἴδωλον.
So that the eye Th is not seen; for it appeared at the intersection of BG, ZTh. Let the same things hold again, and let the eyes B, G be placed above the bisection, at equal distances from the line from the center, ZA. Indeed, I say that B, G are seen, and the right appears left and the left right, and the image is greater than the face, and the image has a greater distance from the mirror.
ἔστω γὰρ ἡ ΒΑ ὄψις ἀνακλωμένη, καὶ ἐπεζεύχθωσαν ἀπὸ τοῦ κέντρου ἐπὶ τὸ Β, Γ αἱ ΖΒ, ΖΓ, καὶ ἐκβεβλήσθω ἡ ΒΑ. ἐπεὶ οὖν διχοτομία ἐστὶ τὸ Ν, μείζων ἐστὶν ἡ ΒΖ τῆς ΒΑ καὶ ἡ Κ γωνία τῆς Ε. ἴση δὲ ἡ Κ τῇ ∠· μείζων ἄρα καὶ ἡ ∠ τῆς Ε. συμπεσοῦνται ἄρα αἱ ΖΒ, ΓΑ ἐκβληθεῖσαι.
For let the visual ray BA be reflected, and let ZB, ZG be joined from the center to B, G, and let BA be extended. Since, therefore, N is the bisection, BZ is greater than BA, and angle K is greater than E. But K is equal to D; therefore, D is also greater than E. Therefore, ZB, GA, being extended, will meet.
συμπιπτέτωσαν κατὰ τὸ Π. διὰ τὰ αὐτὰ δὴ καὶ αἱ ΒΑ, ΖΓ συμπεσοῦνται κατὰ τὸ Θ. ὀφθήσεται ἄρα τὸ μὲν Γ ἐπὶ τοῦ Θ, τὸ δὲ Β ἐπὶ τοῦ Π, καὶ φαίνεται τὰ μὲν δεξιὰ ἀριστερά, τὰ δὲ ἀριστερὰ δεξιά.
Let them meet at P. Indeed, for the same reasons, BA, ZG will also meet at Th. Therefore, G will be seen at Th, and B at P, and the right appears left and the left right.
ἀλλὰ μὴν καὶ μείζων ἡ ΘΠ τῆς ΒΙ παράλληλοι γάρ εἰσιν.
Furthermore, ThP is greater than BI; for they are parallel.
τὸ ἄρα εἴδωλον φαίνεται μεῖζον καὶ μεῖζον ἀπέχον τοῦ ἐνόπτρου· μείζων γὰρ ἡ ΜΑ τῆς ΑΛ.
ἐὰν δὲ ἔξω τῆς διαμέτρου τεθῇ τὰ ὄμματα, τὰ δεξιὰ φαίνεται δεξιὰ καὶ τὰ ἀριστερὰ ἀριστερὰ καὶ τὸ εἴδωλον ἔλασσον τοῦ προσώπου καὶ ἐν τῷ ἀνὰ μέσον τοῦ προσώπου καὶ τοῦ ἐνόπτρου.
Therefore, the image appears greater and having a greater distance from the mirror; for MA is greater than AL. But if the eyes are placed outside the diameter, the right appears right and the left left, and the image is smaller than the face, and in the middle between the face and the mirror.
ἔστω γὰρ ὄμματα τὰ Β, Γ, κέντρον δὲ τὸ Ζ τοῦ ἐνόπτρου, καὶ τῇ διαμέτρῳ πρὸς ὀρθὰς ἔστω ἡ Α Ζ ∠, καὶ ταύτῃ πρὸς ὀρθὰς ἡ ΒΓ, καὶ ἴση τῇ ΒΑ ἔστω ἡ ΑΓ. καὶ ὄψις ἡ Β∠ ἀνακλωμένη ἐπὶ τὸ Γ καὶ διὰ τοῦ κέντρου αἱ ΒΖΚ, ΓΖΕ, καὶ ἀπὸ τῶν Ε, Κ ἡ ΚΕ ἐπεζεύχθω.
For let the eyes be B, G, the center of the mirror Z, and let AZD be perpendicular to the diameter, and let BG be perpendicular to this, and let BA be equal to AG. And let the visual ray BD be reflected to G, and BZK, GZE be through the center, and let KE be joined from E, K.
οὐκοῦν τὸ μὲν Β ἐπὶ τοῦ Κ φαίνεται, τὸ δὲ Γ ἐπὶ τοῦ Ε. τὰ ἄρα δεξιὰ δεξιὰ καὶ τὰ ἀριστερὰ ἀριστερὰ φαίνεται καὶ τὸ ΕΚ εἴδωλον ἔλασσον τοῦ ΒΓ προσώπου· παράλληλος γάρ ἐστιν ἡ ΕΚ τῇ ΒΓ·
Therefore, B is seen at K, and G at E. Therefore, the right appears right and the left left, and the image EK is smaller than the face BG; for EK is parallel to BG.
καὶ ἀνὰ μέσον τοῦ ἐνόπτρου καὶ τοῦ προσώπου φαίνεται τὸ εἴδωλον.
And the image appears between the mirror and the face.
ἀναγομένου δὲ τοῦ προσώπου ἔτι ἔλασσον φαίνεται τὸ εἴδωλον.
And when the face is moved back, the image appears still smaller.
ἔστω γὰρ τὸ ΜΝ πρόσωπον τὸ αὐτὸ τῷ ΒΓ ἀφεστηκὸς ἀπὸ τοῦ ΒΓ κείμενον ὁμοίως.
For let the face MN be the same as BG, standing back from BG and placed similarly.
οὐκοῦν ἡ ἀπὸ τοῦ Μ ἐπὶ τὸ Ζ κέντρον ἐπιζευχθεῖσα καὶ ἐκβληθεῖσα ἀνώτερον πεσεῖται τοῦ Κ ὡς τὸ Λ, ἡ δὲ ἀπὸ τοῦ Ν ἐπὶ τὸ Ζ ἀνώτερον τοῦ Ε ὡς τὸ Θ. φαίνεται ἄρα τὸ ΜΝ ὡς τὸ ΘΛ. καί ἐστιν ἔλασσον τὸ ΘΛ τοῦ ΕΚ καὶ ἔγγιον τοῦ ἐνόπτρου.
Therefore, the line joined from M to the center Z and extended will fall higher than K as L, and that from N to Z higher than E as Th. Therefore, MN appears as ThL. And ThL is smaller than EK and closer to the mirror.