§6.29κθ΄.
29.
Τμήματος δοθέντος τοῦ ἐπὶ τῆς ΑΒ κλάσαι εὐθεῖαν θεῖαν τὴν ΑΓΒ ἐν λόγῳ τῷ δοθέντι.
On a given segment on ΑΒ, to inflect the straight line ΑΓΒ in a given ratio.
γεγονέτω, καὶ διήχθω ἀπὸ τοῦ Γ ἐφαπτομένη ἡ Γ∠ ὡς ἄρα τὸ ἀπὸ ΑΓ πρὸς τὸ ἀπὸ ΒΓ, οὕτως ἡ Α∠ πρὸς ∠ Β. λόγος δὲ τοῦ ἀπὸ ΑΓ πρὸς τὸ ἀπὸ ΓΒ, δοθείς·
Let it have been done, and let the tangent Γ∠ be drawn from Γ; therefore, as the square on ΑΓ is to the square on ΒΓ, so is Α∠ to ∠Β.
ὥστε καὶ ὁ τῆς Α∠ πρὸς τὴν Β∠ δοθείς.
And the ratio of the square on ΑΓ to the square on ΓΒ is given; so that the ratio of Α∠ to Β∠ is also given.
καί ἐστιν δύο δοθέντα τὰ Α, Β· δοθὲν ἄρα ἐστὶν τὸ ∠· ὥστε καὶ τὸ Γ δοθέν.
And there are two given (points) Α, Β; therefore ∠ is given; so that Γ is also given.
Συντεθήσεται δὴ τὸ πρόβλημα οὕτως. ἔστω τὸ μὲν τμῆμα τὸ ΑΒΓ, ὁ δὲ λόγος ὁ τῆς πρὸς τὴν Ζ, καὶ πεποιήσθω, ὡς τὸ ἀπὸ Ε πρὸς τὸ ἀπὸ Ζ, οὕτως ἡ Α∠ πρὸς τὴν ∠Β, καὶ ἤχθω ἐφαπτομένη ἡ ∠Γ, καὶ ἐπεζεύχθωσαν αἱ ΑΓ, ΓΒ. λέγω, ὅτι αἱ ΑΓ, ΓΒ ποιοῦσι τὸ πρόβλημα.
The problem will be synthesized thus: let the segment be ΑΒΓ, and the ratio be that of [Ε] to Ζ, and let it be made that, as the square on Ε is to the square on Ζ, so is Α∠ to ∠Β, and let the tangent ∠Γ be drawn, and let ΑΓ, ΓΒ be joined. I say that ΑΓ, ΓΒ solve the problem.
ἐπεὶ γάρ ἐστιν, ὡς τὸ ἀπὸ Ε πρὸς τὸ ἀπὸ Ζ, οὕτως ἡ Α∠ πρὸς τὴν ∠Β, ὡς δὲ ἡ Α∠ πρὸς τὴν ∠Β, οὕτως τὸ ἀπὸ ΑΓ πρὸς τὸ ἀπὸ ΓΒ διὰ τὸ ἔφάπτεσθαι τὴν ΓΔ, καὶ ὡς ἄρα τὸ ἀπὸ Ε πρὸς τὸ ἀπὸ Ζ, οὕτως τὸ ἀπὸ ΑΓ πρὸς τὸ ἀπὸ ΓΒ ὥστε καί, ὡς ἡ πρὸς τὴν Ζ, οὕτως ἡ ΑΓ πρὸς τὴν ΓΒ. ἡ ΑΓΒ ἄρα ποιεῖ τὸ πρόβλημα.
For since, as the square on Ε is to the square on Ζ, so is Α∠ to ∠Β, and as Α∠ is to ∠Β, so is the square on ΑΓ to the square on ΓΒ because of ΓΔ being tangent, therefore also as the square on Ε is to the square on Ζ, so is the square on ΑΓ to the square on ΓΒ; so that also, as [Ε] is to Ζ, so is ΑΓ to ΓΒ. Therefore the inflected line ΑΓΒ solves the problem.
§6.30λ΄.
30.
Κύκλος, οὗ διάμετρος ἡ ΑΒ, καὶ ἀπὸ τυχόντος ἐπ᾿ αὐτὴν κάθετος ἡ ∠Ε, διήχθω ἡ ∠Ζ, ἐπεζεύχθω ἡ ΕΖ καὶ ἐκβεβλήσθω, καί, καθ᾿ ὃ συμπίπτει τῇ διαμέτρῳ, ἔστω τὸ Η·
Let there be a circle whose diameter is ΑΒ, and from any (point) on it let a perpendicular to it (the diameter) be ∠Ε, and let ∠Ζ be drawn, and let ΕΖ be joined and produced, and let the point where it meets the diameter be Η; (to prove) that, as ΑΗ is to ΗΒ, so is ΑΘ to ΘΒ.
ὅτι ἐστίν, ὡς ἡ ΑΗ πρὸς τὴν ΗΒ, οὕτως ἡ ΑΘ πρὸς τὴν ΘΒ.
ἐπεζεύχθωσαν αἱ ∠Α, ΑΕ, ΑΖ.
ἐπεὶ οὖν ἐπὶ διάμετρον κάθετος ἡ ∠Ε, ἴση ἐστὶν ἡ ὑπὸ ∠ΑΒ τῇ ὑπὸ ΒΑΕ. ἀλλ᾿ ἡ ὑπὸ ∠ΑΒ τῇ ἐν τῷ αὐτῷ τμήματι ἴση ἐστὶν τῇ ὑπὸ ΘΖΒ, ἡ δὲ ὑπὸ ΒΑΕ ἴση ἐστὶν τῇ ἐκτὸς τετραπλεύρου τῇ ὑπὸ ΒΖΗ καὶ ἡ ὑπὸ ΘΖΒ ἄρα γωνία ἴση ἐστὶν τῇ ὑπὸ ΒΖΗ. καί ἐστιν ὀρθὴ ἡ ὑπὸ ΑΖΒ γωνία· διὰ δὴ τὸ λῆμμα γίνεται, ὡς ἡ ΑΗ πρὸς τὴν ΗΒ, οὕτως ἡ Α πρὸς τὴν ΒΘ.
Let ∠Α, ΑΕ, ΑΖ be joined. Since, therefore, ∠Ε is a perpendicular to the diameter, angle ∠ΑΒ is equal to angle ΒΑΕ. But angle ∠ΑΒ is equal to angle ΘΖΒ in the same segment, and angle ΒΑΕ is equal to angle ΒΖΗ, which is exterior to the quadrilateral; therefore also angle ΘΖΒ is equal to angle ΒΖΗ. And angle ΑΖΒ is a right angle; therefore by the lemma, as ΑΗ is to ΗΒ, so is Α[Θ] to [Θ]Β.
§6.31λα΄.
31.
Ἡμικύκλιον τὸ ἐπὶ τῆς ΑΒ, καὶ ἀπὸ τῶν Α,Β σημείων τῇ ΑΒ πρὸς ὀρθὰς γωνίας εὐθεῖαι γραμμαὶ ἤχθωσαν αἱ ΒΔ, ΑΕ, καὶ ἤχθω τυχοῦσα ἡ ΔΕ, καὶ ἀπὸ τοῦ τῇ ∠Ε πρὸς ὀρθὰς γωνίας εὐθεῖα γραμμὴ ἡ ΖΗ συμπιπτέτω τῇ ΑΒ κατὰ τὸ Η ὅτι τὸ ὑπὸ τῶν ΑΕ, ΒΔ ἴσον ἐστὶν τῷ ὑπὸ τῶν ΑΗΒ.
ὅτι ἄρα ἐστίν, ὡς ἡ ΕΑ πρὸς τὴν ΑΗ, οὕτως ἡ ΗΒ πρὸς τὴν Β∠.
Let there be a semicircle on ΑΒ, and from points Α, Β let straight lines ΒΔ, ΑΕ be drawn at right angles to ΑΒ, and let an arbitrary straight line ΔΕ be drawn, and from [Ζ] let a straight line ΖΗ at right angles to ∠Ε meet ΑΒ at Η; (to prove) that the rectangle contained by ΑΕ, ΒΔ is equal to the rectangle contained by ΑΗ, ΗΒ. That is, as ΕΑ is to ΑΗ, so is ΗΒ to Β∠.
περὶ ἴσας γωνίας ἀνάλογόν εἰσιν αἱ πλευραί· ὅτι ἄρα ἴση ἐστὶν ἡ ὑπὸ τῶν ΑΗΕ γωνία τῇ ὑπὸ τῶν ΒΔΗ γωνίᾳ.
The sides about equal angles are proportional; therefore angle ΑΗΕ is equal to angle ΒΔΗ.
ἀλλὰ ἡ μὲν ὑπὸ ΑΗΕ ἴση ἐστὶν ἐν τῷ αὐτῷ τμήματι τῇ ὑπὸ ΑΖΕ, ἡ δὲ ὑπὸ Β∠Η πάλιν ἐν τῷ αὐτῷ τμήματι τῇ ὑπὸ ΒΖΗ ὅτι ἄρα ἴση ἐστὶν ἡ ὑπὸ ΑΖΕ γωνία τῇ ὑπὸ ΒΖΗ γωνίᾳ.
But angle ΑΗΕ is equal to angle ΑΖΕ in the same segment, and angle Β∠Η also in the same segment is equal to angle ΒΖΗ; therefore angle ΑΖΕ is equal to angle ΒΖΗ.
ἔστιν δέ· ὀρθὴ γάρ ἐστιν ἑκατέρα τῶν ὑπὸ ΑΖΒ, ΕΖΗ γωνιῶν.
And indeed it is so; for each of the angles ΑΖΒ, ΕΖΗ is a right angle.
§6.32τῷ ΑΒΓ τριγώνῳ, κοινοῦ δ᾿ ἀφαιρουμένου τοῦ ΑΒΕ λοιπὸν τὸ ∠ΑΕ λοιπῷ τῷ ΑΓE ἐστιν ἴσον καί ἐστιν ἐπὶ τῆς αὐτῆς βάσεως.
In the triangle ΑΒΓ, when the common ΑΒΕ is subtracted, the remaining ∠ΑΕ is equal to the remaining ΑΓΕ, and they are on the same base.