§19## ιθ΄.
## 19.
Ἐν τοῖς ἐπιπέδοις ἐνόπτροις τὰ δεξιὰ ἀριστερὰ φαίνεται καὶ τὰ ἀριστερὰ δεξιὰ καὶ τὸ εἴδωλον ἴσον τῷ ὁρωμένῳ, καὶ τὸ ἀπόστημα τὸ ἀπὸ τοῦ ἐνόπτρου ἴσον ἐστίν.
In plane mirrors, right appears left and left right, and the image is equal to the object seen, and the distance from the mirror is equal.
ἔστω ἐπίπεδον ἔνοπτρον τὸ ΑΓ, ὄμμα δὲ τὸ Β, ὄψεις δὲ αἱ ΒΑ, ΒΓ ἀνακλώμεναι ἐπὶ τὰ Ε, ∠, ὁρώμενον δὲ ἔστω τὸ Ε ∠, καὶ ἀπὸ τῶν Ε, ∠ ἐπὶ τὸ ἔνοπτρον κάθετοι ἤχθωσαν αἱ Ε Ζ, ∠Θ καὶ ἐκβεβλήσθωσαν, ἐκβεβλήσθωσαν δὲ καὶ αἱ ΒΓ, ΒΑ ὄψεις καὶ συμπιπτέτωσαν ταῖς καθέτοις κατὰ τὰ Κ, Λ, καὶ ἐπεζεύχθω ἡ ΛΚ. οὐκοῦν φαίνεται τὸ μὲν Ε ἐπὶ τοῦ Κ, τὸ δὲ ∠ ἐπὶ τοῦ Λ· τοῦτο γὰρ προεδείχθη.
Let the plane mirror be AG, the eye B, and the visual rays BA, BG reflected to E, D, and let the object seen be ED. And from E, D let perpendiculars EZ, DTh be drawn to the mirror, and let them be extended; and let the visual rays BG, BA also be extended, and let them meet the perpendiculars at K, L, and let LK be joined. Therefore, E appears at K, and D at L; for this was shown before.
τὰ ἄρα ἀριστερ δεξιὰ φαίνεται καὶ τὰ δεξιὰ ἀριστερά.
Therefore, left appears right, and right left.
καὶ ἐπεὶ ἴσ ἐστὶν ἡ ὑπὸ τῶν ΚΓΖ γωνία τῇ ὑπὸ τῶν ΖΙΕ, κ εἰσιν ὀρθαὶ αἱ πρὸς τῷ Ζ, ἴση ἂν εἴη καὶ ἡ ΖΚ τ ΖΕ. διὰ τὰ αὐτὰ καὶ ἡ ∠Θ τῇ ΘΑ. ἴσον ἄρα τὸ ἀπόστημα, ὃ ἀπέχει ἀπὸ τοῦ ἐνόπτρου τὸ Ε∠, τῷ, ὃ ἀπέχει τὸ εἴδωλον τὸ Κ Λ. καὶ ἴσον τὸ ὁρώμενον τὸ Ε∠ τῷ εἰδώλῳ τῷ Κ Λ διὰ τὸ ἴσην εἶναι τὴν μὲν ΕΖ τῇ ΖΚ, τὴν δὲ ∠Θ τῇ ΘΛ, κοινὴν δὲ καὶ πρὸς ὀρθὰς τὴν ΘΖ.
And since the angle KGZ is equal to ZGE, and the angles at Z are right, ZK would also be equal to ZE. For the same reasons, also DTh to ThL. Therefore, the distance which the object ED is distant from the mirror is equal to that which the image KL is distant. And the object seen ED is equal to the image KL, because EZ is equal to ZK, and DTh to ThL, and ThZ is common and perpendicular.
§20## κ΄.
## 20.
Ἐν τοῖς κυρτοῖς ἐνόπτροις τὰ ἀριστερὰ δεξιὰ φαίνεται καὶ τὰ δεξιὰ ἀριστερά, καὶ τὸ ἀπόστημα ἀπὸ τοῦ ἐνόπτρου τὸ εἴδωλον ἔλασσον ἔχει.
In convex mirrors, left appears right and right left, and the image has a smaller distance from the mirror.
ἔστω ἔνοπτρον κυρτὸν τὸ ΑΓ, κέντρον δὲ τῆς σφαίρας τὸ Θ, ὄμμα δὲ τὸ Β, ὄψεῖς δὲ αἱ ΒΑ, ΒΓ ἀνακλώμεναι ἐπὶ τὰ ∠, Ε, ὁρώμενον δὲ τὸ ∠Ε, καὶ ἀπὸ τοῦ Θ κέντρου ἤχθωσαν ἐπὶ τὰ ∠, Ε αἱ Θ∠, ΘΕ, καὶ ἐκβεβλήσθωσαν αἱ ὄψεις ἐπὶ τὰ Ζ, Η, καὶ ἐπεζεύχθω τὸ ΖΗ εἴδωλον.
Let the convex mirror be AG, the center of the sphere Th, the eye B, and the visual rays BA, BG reflected to D, E, and let the object seen be DE, and from the center Th let ThD, ThE be drawn to D, E, and let the visual rays be extended to Z, H, and let the image ZH be joined.
οὐκοῦν τὸ μὲν ∠ φαίνεται ἐπὶ τοῦ Η, τὸ δὲ Ε ἐπὶ τοῦ Ζ. τὰ ἄρα δεξιὰ ἀριστερὰ φαίνεται καὶ τὰ ἀριστερὰ δεξιά.
Therefore, D appears at H, and E at Z. Therefore, right appears left, and left right.
λέγω, ὅτι μείζων ἐστὶν ἡ ΕΛ τῆς ΛΖ. ἤχθω γὰρ διὰ τοῦ Α ἐφαπτομένη τῆς περιφερείας ἡ ΡΑΚ. ἐπεὶ οὖν αἱ ΒΑ, ΑΕ πρὸς τὴν περιφέρειαν ἴσας ποιοῦσι γωνίας διὰ τὴν ἀνάκλασιν, ἐφάπτεται δὲ ἡ ΚΑΡ δίχα ἂν εἴη τετμημένη ἡ ὑπὸ τῶν ΕΑΖ γωνία.
I say that EL is greater than LZ. For let RAK be drawn through A, tangent to the circumference. Since, therefore, BA, AE make equal angles with the circumference because of reflection, and KAP is tangent, the angle EAZ would be bisected.
καὶ ἀμβλεῖά ἐστιν ἡ Κ γωνία· μείζων ἄρα ἡ ΕΚ τῆς ΚΖ πολλῷ μᾶλλον ἡ ΕΛ τῆς ΛΖ. ἔλασσον ἄρα ἀπέχει τὸ εἴδωλον τὸ ΖΗ ἀπὸ τοῦ ἐνόπτρου, μεῖζον δὲ τὸ ὁρώμενον τὸ Ε∠.
And the angle K is obtuse; therefore, EK is greater than KZ, and much more so is EL than LZ. Therefore, the image ZH is less distant from the mirror, and the object seen ED is more distant.
§21## κα΄.
## 21.
Ἐν τοῖς κυρτοῖς ἐνόπτροις τὸ εἴδωλον ἔλασσόν ἐστι τῶν ὁρωμένων.
In convex mirrors, the image is smaller than the objects seen.
ἔστω γὰρ κυρτὸν ἔνοπτρον τὸ ΑΟΓ, ὄμμα δὲ τὸ Β, ὄψεις δὲ ἀνακλώμεναι αἱ ΒΑ, ΒΓ ἐπὶ τὰ ∠, Ε. οὐκοῦν ἀπὸ τοῦ κυρτοῦ ἐνόπτρου θεωρεῖται τὸ Ε∠ ἐν γωνίᾳ τῇ ὑπὸ ΑΒΓ. παρακείσθω δὴ ἔνοπτρον ἐπίπεδον τὸ ΑΓ ἁπτόμενον τῶν ὄψεων κατὰ τὰ Α, Γ. οὐκοῦν ἡ ὄψις ἡ μέλλουσα ἰδεῖν τὸ Ε ἀπὸ τοῦ ἐπιπέδου ἐνόπτρου οὐκ ἔστιν ἡ ΒΑΕ· οὐ γὰρ ποιεῖ γωνίας ἴσας πρὸς τῷ ἐπιπέδῳ ἐνόπτρῳ.
For let the convex mirror be AOG, the eye B, and the visual rays BA, BG reflected to D, E. Therefore, from the convex mirror, ED is viewed under the angle ABG. Now let there be placed alongside a plane mirror AG touching the visual rays at A, G. Therefore, the visual ray which is going to see E from the plane mirror is not BAE; for it does not make equal angles with the plane mirror.
οὐδὲ μὴν κλασθήσεται μεταξὺ τῶν Α, Γ. κεκλάσθω γάρ, εἰ δυνατόν, καὶ ἔστω ἡ ΒΖΕ ὄψις.
Nor indeed will it be reflected between A and G. For let it be reflected, if possible, and let it be the visual ray BZE.
ἴση ἄρα ἡ Η γωνία τῇ Θ διὰ τὴν ἀνάκλασιν.
Therefore, the angle H is equal to Th because of reflection.
ἡ δὲ Θ μείζων τῆς ΝΙ, ἡ δὲ Μ τῆς Η· ὥστε καὶ ἡ Μ τῆς ΝΙ μείζων ἐστίν· ὅπερ ἀδύνατον.
But Th is greater than NI, and M is greater than H; so that M is also greater than NI; which is impossible.
αὐτὴ γὰρ ἡ Ι μείζων τῆς Μ ἐστιν· ἴση γάρ ἐστιν ὅλῃ τῇ πρὸς τῇ περιφερείᾳ.
For I itself is greater than M; for it is equal to the whole angle at the circumference.
ἐκτὸς ἄρα ἀνακλασθήσεται τοῦ Α. κεκλάσθω καὶ ἔστω ἡ ΒΚΕ. ὁμοίως δὲ καὶ ἡ ΒΛ∠ πεσεῖται ἐκτός.
Therefore, it will be reflected outside of A. Let it be reflected and let it be BKE. Similarly, BLD will also fall outside.
τὸ ἄρα Ε∠ ὑπὸ μείζονος γωνίας θεωρεῖται ἀπὸ τοῦ ἐπιπέδου ἐνόπτρου τῆς περιεχομένης ὑπὸ ΚΒΛ ἤπερ ἀπὸ τοῦ κυρτοῦ.
Therefore, ED is viewed under a greater angle from the plane mirror, namely that contained by KBL, than from the convex mirror.
ἴσον δὲ ἐδείχθη φαινόμενον ἐν τῷ ἐπιπέδῳ ἐνόπτρῳ.
And it was shown to appear equal in the plane mirror.
φανερὸν οὖν, ὅτι ἀπὸ τοῦ κυρτοῦ ἐνόπτρου τὸ εἴδωλον ἔλασσον φαίνεται τοῦ ὁρωμένου.
It is clear, therefore, that from the convex mirror the image appears smaller than the object seen.