§3## γ΄.
## 3.
Πρὸς ὁποῖον ἂν τῶν ἐνόπτρων προσπίπτουσα ὄψις ἀνίσους ποιῇ γωνίας, οὔτε διʼ ἑαυτῆς ἀνακλασθήσεται οὔτε ἐπὶ τῆς ἐλάσσονος γωνίας.
Onto whatever kind of mirror a falling visual ray makes unequal angles, it will be reflected neither along itself nor towards the side of the smaller angle.
ἔστω ἐπίπεδον ἔνοπτρον τὸ ΑΚΗΓ, ὄψις δὲ ἡ ΒΚ προσπιπτέτω μείζονα ποιοῦσα γωνίαν τὴν Ζ τῆς Θ, Λ. λέγω, ὅτι ἡ ΒΚ ἀνακλωμένη οὔτε αὐτὴ διʼ ἑαυτῆς ἀνακλασθήσεται οὔτε ἐπὶ τὴν Θ, Λ γωνίαν.
Let the plane mirror be AKHG, and let the visual ray BK fall making angle Z greater than Θ, Λ. I say that the reflected BK will be reflected neither along itself nor towards the angle Θ, Λ.
εἰ μὲν γὰρ ἥξει ἐπὶ τὸ Β, ἔσται ἡ Ζ γωνία τῇ Θ, Λ ἴση· ὅπερ ἄτοπον· ὑπόκειται γὰρ μείζων.
For if it returns to B, angle Z will be equal to Θ, Λ; which is absurd; for it is assumed to be greater.
εἰ δὲ διὰ τοῦ ∠, ἴση ἔσται ἡ Ζ γωνία τῇ Θ· ἔστι δὲ μείζων.
But if it goes through ∠, angle Z will be equal to Θ; but it is greater.
ἡ ἄρα ΒΚ ἀνακλασθήσεται ἐπὶ τὴν μείζονα γωνίαν τὴν Ζ· δυνατὸν γὰρ ἀπὸ τῆς μείζονος τῇ ἐλάσσονι ἴσην ἀφαιρεθῆναι.
Therefore, BK will be reflected towards the greater angle Z; for it is possible for an equal to be subtracted from the greater for the less.
ἔστι δὲ ἡ αὐτὴ ἀπόδειξις ἐπὶ τῶν κυρτῶν καὶ κοίλων.
And the same proof holds for convex and concave mirrors.
§4## δ΄.
## 4.
Αἱ ὄψεις ἐπὶ τῶν ἐπιπέδων ἐνόπτρων καὶ κυρτῶν ἀνακλώμεναι οὔτε συμπεσοῦνται ἀλλήλαις οὔτε παράλληλοι ἔσονται.
Visual rays reflected on plane and convex mirrors will neither meet each other nor be parallel.
ἔστω ἐπίπεδον ἔνοπτρον τὸ ΑΓ, ὄμμα δὲ τὸ Β, ὄψεις δὲ ἀνακλώμεναι αἱ ΒΓ∠, ΒΑΕ. λέγω, ὅτι αἱ Γ∠, ΑΕ οὔτε παράλληλοί εἰσιν οὔτε συμπεσοῦνται ἐπὶ τὰ ∠, Ε. ἐπεὶ γὰρ ἴση ἐστὶν ἡ γωνία τῇ Θ, ἡ δὲ Κ τῇ Μ, μείζων δὲ ἡ Ζ τῆς Κ διὸ τὸ ἐκτὸς εἶναι ἐν τῷ ΒΑΓ τριγώνῳ, μείζων ἂν εἴη καὶ ἡ Θ τῆς Μ. οὐκ ἄρα παράλληλος ἡ Γ∠ τῇ ΑΕ ἐστιν, οὐδὲ συμπίπτουσιν ἐπὶ τὰ Ε, ∠.
Let the plane mirror be AC, the eye B, and the reflected visual rays BCD, BAE. I say that GD, AE are neither parallel nor will they meet towards D, E. For since angle [Z] is equal to Θ, and K to M, and Z is greater than K because of its being external in triangle BAG, then Θ would also be greater than M. Therefore, GD is not parallel to AE, nor do they meet towards E, D.
ἔστω πάλιν κυρτὸν ἔνοπτρον τὸ ΑΖΓ, ὄμμα δὲ τὸ Β, ὄψεις δὲ ἀνακλώμεναι αἱ ΒΖ∠, ΒΗΕ. λέγω, ὅτι αἱ Ζ∠, ΕΗ οὔτε παράλληλοί εἰσιν οὔτε συμπεσοῦνται ἐπὶ τὰ Ε, ∠.
Now let the convex mirror be AZG, the eye B, and the reflected visual rays BZD, BHE. I say that ZD, EH are neither parallel nor will they meet towards E, D.
ἐπεζεύχθω γὰρ ἡ ΗΖ εὐθεῖα καὶ ἐκβεβλήσθω ἐφʼ ἑκάτερα.
For let the straight line HZ be joined and extended in both directions.
ἐπεὶ ἴση ἐστὶν ἡ Κ, Θ τῇ Λ διὰ τὸ ἐν ἴσαις ἀνακλᾶσθαι γωνίαις, εἴη ἂν μείζων ἡ Λ, Μ τῆς Κ. ἡ δὲ Κ τῆς Ν, Ξ ἐστι μείζων, ἡ δὲ Ν, Ξ τῆς Ο, Π μείζων· αὐτὴ γὰρ ἡ Ξ ἴση ἐστὶ τῇ Ο, Π·
Since K, Θ is equal to Λ because of being reflected at equal angles, Λ, M would be greater than K. And K is greater than N, Ξ, and N, Ξ is greater than O, Π; for Ξ itself is equal to O, Π; therefore, Λ, M is greater than O, Π.
μείζων ἄρα ἡ Λ, Μ τῆς Ο, Π. πολλῷ ἄρα ἡ Λ, Μ τῆς Ο μείζων ἐστίν.
Therefore, Λ, M is much greater than O.
οὐκ ἄρα συμπεσοῦνται αἱ Ζ∠, ΗΕ εὐθεῖαι οὐδὲ παράλληλοί εἰσιν.
Therefore, the straight lines ZD, HE will not meet, nor are they parallel.
§5## ε΄.
## 5.
Ἐν τοῖς κοίλοις ἐνόπτροις ἐὰν ἢ ἐπὶ τὸ κέντρον ἢ ἐπὶ τῆς περιφερείας ἢ ἐκτὸς τῆς περιφερείας θῇς τὸ ὄμμα, τουτέστι μεταξὺ τοῦ κέντρου καὶ τῆς περιφερείας, αἱ ὄψεις ἀνακλώμεναι συμπεσοῦνται.
In concave mirrors, if you place the eye either at the center, or on the circumference, or outside the circumference—that is, between the center and the circumference—the reflected visual rays will meet.
ἔστω κοῖλον ἔνοπτρον τὸ ΑΓ∠, κέντρον δὲ τῆς σφαίρας τὸ Β, καὶ κείσθω τὸ ὄμμα ἐπὶ τοῦ Β, καὶ προσπιπτέτωσαν ἀπὸ τοῦ Β ὄψεις πρὸς τὴν περιφέρειαν αἱ ΒΑ, ΒΓ, Β∠.
Let the concave mirror be AGD, the center of the sphere B, and let the eye be placed at B, and from B let the visual rays BA, BG, BD fall on the circumference.
ἴσαι ἄρα εἰσὶν αἱ πρὸς τοῖς σημείοις τοῖς Α, ∠, Γ γωνίαι· ἡμικυκλίου γάρ εἰσιν.
Therefore, the angles at points A, D, G are equal; for they are angles of a semicircle.
αἱ ἄρα ὄψεις ἀνακλώμεναι διʼ ἑαυτῶν ἀνακλασθήσονται αἱ ΒΑ, ΒΓ, Β∠ τοῦτο γὰρ δέδεικται.
Therefore, the reflected visual rays BA, BG, BD will be reflected along themselves; for this has been shown.
ὥστε συμπεσοῦνται κατὰ τὸ Β.
ἔστω πάλιν κοῖλον ἔνοπτρον τὸ ΑΓΒ, ὄμμα δὲ τὸ Β, κείσθω δὲ ἐπὶ τῆς περιφερείας αὐτοῦ, καὶ ἀπὸ τοῦ Β προσπιπτέτωσαν ὄψεις αἱ ΒΓ, ΒΑ ἀνακλώμεναι ἐπὶ τὰ ∠, Ε σημεῖα.
Thus they will meet at B. Let again the concave mirror be AGB, the eye B, and let it be placed on its circumference, and from B let the visual rays BG, BA fall and be reflected to points D, E.
ἐπεὶ μεῖζον τὸ ΑΓΒ τμῆμα τοῦ ΒΓ τμήματος, μείζων ἡ Ζ γωνία τῆς γωνίας.
Since the segment AGB is greater than the segment BG, angle Z is greater than angle [Θ].
καὶ ἡ ἄρα τῆς Κ μείζων.
Therefore, [angle Λ] is also greater than K.
αἱ ἄρα Ζ, τῶν Θ, Κ μείζους εἰσίν.
Therefore, Z, [Λ] are greater than Θ, K.
λοιπὴ ἄρα ἡ Λ τῆς Μ ἐλάσσων·
Therefore, the remaining Λ is less than M; therefore, much less than N.
πολλῷ μᾶλλον ἄρα τῆς Ν. συμπεσοῦνται ἄρα αἱ Γ∠, ΑΕ κατὰ τὸ Ξ ὁμοίως δειχθήσεται, κἂν ἐκτὸς τῆς περιφερείας πίπτῃ τὸ ὄμμα, ὡς ἐπὶ τοῦ ἑξῆς θεωρήματος.
Therefore, GD, AE will meet at X. It will be shown similarly also if the eye falls outside the circumference, as in the next theorem.