§prΌψιν εἶναι εὐθεῖαν, ἧς τὰ μέσα πάντα τοῖς ἄκροις ἐπιπροσθεῖ.
That the visual ray is a straight line, of which all the intermediate parts block the extremities.
Τὰ ὁρώμενα ἅπαντα καθʼ εὐθείας ὁρᾶσθαι.
That all seen objects are seen along straight lines.
Ἐνόπτρου τεθέντος ἐν ἐπιπέδῳ καὶ θεωρουμένου τινὸς ὕψους, ὃ πρὸς ὀρθάς ἐστι τῷ ἐπιπέδῳ, γίγνονται ἀνάλογον, ὡς ἡ μεταξὺ τοῦ ἐνόπτρου καὶ τοῦ θεωροῦντος εὐθεῖα πρὸς τὴν μεταξὺ τοῦ ἐνόπτρου καὶ τοῦ πρὸς ὀρθὰς ὕψους, οὕτω τὸ τοῦ θεωροῦντος ὕψος πρὸς τὸ πρὸς ὀρθὰς τῷ ἐπιπέδῳ ὕψος.
When a mirror is placed on a plane and some height is observed which is perpendicular to the plane, there arises a proportion: as the straight line between the mirror and the observer is to the straight line between the mirror and the perpendicular height, so is the height of the observer to the height perpendicular to the plane.
Ἐν τοῖς ἐπιπέδοις ἐνόπτροις τοῦ τόπου καταληφθέντος, ἐφʼ ὃν ἡ κάθετος πίπτει ἀπὸ τοῦ ὁρωμένου, οὐκέτι ὁρᾶται τὸ ὁρώμενον.
In plane mirrors, if the place is covered upon which the perpendicular falls from the object seen, the object seen is no longer seen.
Καὶ ἐν τοῖς κυρτοῖς ἐνόπτροις καταληφθέντος τοῦ τόπου, διʼ οὗ ἀπὸ τοῦ ὁρωμένου εἰς τὸ κέντρον ἄγεται τῆς σφαίρας, οὐκέτι ὁρᾶται τὸ ὁρώμενον.
And in convex mirrors, if the place is covered through which the line is drawn from the object seen to the center of the sphere, the object seen is no longer seen.
τὸ δʼ αὐτὸ καὶ ἐν τοῖς κοίλοις συμβαίνει.
And the same happens also in concave mirrors.
Ἐὰν εἰς ἀγγεῖον ἐμβληθῇ τι καὶ λάβῃ ἀπόστημα ὡς μηκέτι ὁρᾶσθαι, τοῦ αὐτοῦ ἀποστήματος ὄντος ἐὰν ὕδωρ ἐγχυθῇ, ὀφθήσεται τὸ ἐμβληθέν.
If something is cast into a vessel and is at such a distance as to be no longer seen, if water is poured in, the distance remaining the same, the object cast in will be seen.
§1## α΄.
## 1.
Ἀπὸ τῶν ἐπιπέδων ἐνόπτρων καὶ κυρτῶν καὶ κοίλων αἱ ὄψεις ἐν ἴσαις γωνίαις ἀνακλῶνται.
From plane, convex, and concave mirrors, visual rays are reflected at equal angles.
ἔστω ὄμμα τὸ Β, ἔνοπτρον ἐπίπεδον τὸ ΑΓ. ὄψις δʼ ἀπὸ τοῦ ὄμματος φερέσθω ἡ ΒΚ καὶ ἀνακεκλάσθω ἐπὶ τὸ ∠.
Let the eye be B, and the plane mirror AC. Let the visual ray BK proceed from the eye and be reflected to ∠.
φημὶ δὴ τὴν Ε γωνίαν ἴσην εἶναι τῇ Ζ. ἤχθωσαν κάθετοι ἐπὶ τὸ ἔνοπτρον αἱ ΒΓ, ∠Α. οὐκοῦν ἐστιν, ὡς ἡ ΒΓ πρὸς ΓΚ, ἡ ∠Α πρὸς ΑΚ· τοῦτο γὰρ ἐν τοῖς ὅροις ὑπέκειτο·
I say indeed that angle E is equal to Z. Let perpendiculars BC, ∠A be drawn to the mirror. Therefore it is: as BC is to CK, so is ∠A to AK; for this was assumed in the definitions.
ὅμοιον ἄρα τὸ ΒΓΚ τρί γωνον τῷ ∠ΑΚ τριγώνῳ.
Therefore, triangle BCK is similar to triangle ∠AK.
ἴση ἄρα ἡ Ε γωνία τῇ γωνίᾳ· τὰ γὰρ ὅμοια τρίγωνα ἰσογώνιά ἐστιν.
Therefore, angle E is equal to the angle [Z]; for similar triangles are equiangular.
ἔστω δὴ κυρτὸν ἔνοπτρον τὸ ΑΚΓ, ὄψις δὲ ἡ ΒΚ ἀνακλωμένη ἐπὶ τὸ ∠.
Now let the convex mirror be AKG, and the visual ray BK reflected to ∠.
λέγω, ὅτι ἴδη ἐστὶν ἡ Ε, Θ γωνία τῇ Ζ, Λ. παρέθηκα ἐπίπεδον ἔνοπτρον τὸ ΝΜ·
I say that angle E, Θ is equal to Z, Λ. Let the plane mirror NM be applied.
ἴση ἄρα ἐστὶν ἡ Ε γωνία τῇ Ζ. ἀλλὰ καὶ ἡ Θ τῇ Λ· ἐφάπτεται γὰρ ἡ ΜΝ. ὅλη ἄρα ἡ Ε, Θ ὅλῃ τῇ Λ, Ζ ἐστιν ἴση.
Therefore, angle E is equal to Z. But also Θ is equal to Λ; for MN is tangent. Therefore, the whole E, Θ is equal to the whole Λ, Z.
ἔστω δὴ πάλιν κοῖλον ἔνοπτρον τὸ ΑΚΓ, ὄψις δὲ ἡ ΒΚ ἀνακλωμένη ἐπὶ τὸ ∠.
Let again the concave mirror be AKG, and the visual ray BK reflected to ∠.
λέγω, ὅτι ἡ Ε γωνία ἴση ἐστὶ τῇ Ζ. παρατεθέντος γὰρ ἐπιπέδου ἐνόπτρου ἴση γίγνεται ἡ Θ, Ε γωνία τῇ Ζ, Λ· ἴση δὲ καὶ ἡ Θ τῇ Λ· λοιπὴ ἄρα ἡ Ε τῇ Ζ ἴση ἔσται.
I say that angle E is equal to Z. For when a plane mirror is applied, angle Θ, E becomes equal to Z, Λ; and Θ is also equal to Λ; therefore, the remaining E will be equal to Z.
§2## β΄.
## 2.
Πρὸς ὁποῖον ἂν τῶν ἐνόπτρων προσπέσῃ ὄψις ἴσας ποιοῦσα γωνίας, αὐτὴ διʼ ἑαυτῆς ἀνακλασθήσεται.
Onto whatever kind of mirror a visual ray falls making equal angles, it will be reflected along itself.
ἔστω ἔνοπτρον ἐπίπεδον τὸ ΑΓ, ὄμμα δὲ τὸ Β, ὄψις δὲ ἡ ΒΚ προσπεπτωκέτω ἴσας ποιοῦσα γωνίας τὴν Ε, Ζ τῇ Θ. λέγω, ὅτι ἀνακλωμένη ἡ ΒΚ ἐφʼ ἑαυτῆς ἥξει, τουτέστιν ἐπὶ τὸ Β. μὴ γάρ, ἀλλʼ εἰ δυνατόν, ἡκέτω ἐπὶ τὸ ∠.
Let the plane mirror be AC, the eye B, and let the visual ray BK fall making equal angles, namely the angle E, Z equal to Θ. I say that the reflected BK will return along itself, that is, to B. For if not, if possible, let it return to ∠.
καὶ ἐπειδὴ αἱ ὄψεις ἐν ἴσαις ἀνακλῶνται γωνίαις, ἴση ἐστὶν ἡ Ε γωνία τῇ Θ, ἐδείχθη δὲ καὶ ἡ Ε, Ζ γωνία τῇ Θ ἴση.
And since visual rays are reflected at equal angles, angle E is equal to Θ; but angle E, Z was also shown to be equal to Θ.
καὶ ἡ Ε, Ζ ἄρα γωνία τῇ Ε γωνίᾳ ἔσται ἴση, ἡ μείζων τῇ ἐλάσσονι· ὅπερ ἐστὶν ἀδύνατον.
Therefore, angle E, Z will also be equal to angle E, the greater to the less; which is impossible.
ἡ ἄρα ΒΚ διʼ αὑτῆς ἀνακλασθήσεται.
Therefore, BK will be reflected along itself.
ἡ δʼ αὐτὴ ἀπόδειξις ἁρμόσειεν ἂν ἐπὶ τῶν κυρτῶν καὶ τῶν κοίλων ἐνόπτρων.
And the same proof would apply to convex and concave mirrors.