§6.4δ΄.
4.
Καταγραφὴ ἡ ΑΒΓ∠ΕΖΗΘΚΛ, ἔστω δέ, ὡς τὸ ὑπὸ ΑΖ, ΒΓ πρὸς τὸ ὑπὸ ΑΒ, ΓΖ, οὕτως τὸ ὑπὸ ΑΖ, ∠Ε πρὸς τὸ ὑπὸ Α∠, ΕΖ· ὅτι εὐθεῖά ἐστιν ἡ διὰ τῶν Θ, Η, Ζ σημείων.
Let the figure be ΑΒΓ∠ΕΖΗΘΚΛ, and let, as the product of ΑΖ, ΒΓ is to the product of ΑΒ, ΓΖ, so the product of ΑΖ, ∠Ε is to the product of Α∠, ΕΖ; (to prove) that the (line) through the points Θ, Η, Ζ is a straight line.
ἐπεί ἐστιν, ὡς τὸ ὑπὸ ΑΖ, ΒΓ πρὸς τὸ ὑπὸ ΑΒ, ΓΖ, οὕτως τὸ ὑπὸ ΑΖ, ∠Ε πρὸς τὸ ὑπὸ Α∠, ΕΖ, ἐναλλάξ ἐστιν, ὡς τὸ ὑπὸ ΑΖ, ΒΓ πρὸς τὸ ὑπὸ ΑΖ, ∠Ε, τουτέστιν ὡς ἡ ΒΓ πρὸς τὴν ∠Ε, οὕτως τὸ ὑπὸ ΑΒ, ΓΖ πρὸς τὸ ὑπὸ Α∠, ΕΖ. ἀλλ ὁ μὲν τῆς ΒΓ πρὸς τὴν ∠Ε συνῆπται λόγος, ἐὰν διὰ τοῦ Κ τῇ ΑΖ παράλληλος ἀχθῇ ἡ ΚΜ, ἔκ τε τοῦ τῆς ΒΓ πρὸς ΚΝ καὶ τῆς ΚΝ πρὸς ΚΜ καὶ ἔτι τοῦ τῆς ΚΜ πρὸς ∠Ε, ὁ δὲ τοῦ ὑπὸ ΑΒ, ΓΖ πρὸς τὸ ὑπὸ Α∠, ΕΖ συνῆπται ἔκ τε τοῦ τῆς ΒΑ πρὸς Α∠ καὶ τοῦ τῆς ΓΖ πρὸς τὴν ΖΕ. κοινὸς ἐκκεκρούσθω ὁ τῆς ΒΑ πρὸς Α∠ ὁ αὐτὸς ὢν τῷ τῆς ΝΚ πρὸς ΚΜ· λοιπὸν ἄρα ὁ τῆς ΓΖ πρὸς τὴν ΖΕ συνῆπται ἔκ τε τοῦ τῆς ΒΓ πρὸς τὴν ΚΝ, τουτέστιν τοῦ τῆς ΘΓ πρὸς τὴν ΚΘ, καὶ τοῦ τῆς ΚΜ πρὸς τὴν ∠Ε, τουτέστιν τοῦ τῆς ΚΗ πρὸς τὴν ΗΕ. εὐθεῖα ἄρα ἡ διὰ τῶν Θ, Η, Ζ.
ἐὰν γὰρ διὰ τοῦ Ε τῇ Θ παράλληλον ἀγάγω τὴν ΕΞ, καὶ ἐπιζευχθεῖσα ἡ ΘΗ ἐκβληθῇ ἐπὶ τὸ Ξ, ὁ μὲν τῆς ΚΗ πρὸς τὴν ΗΕ λόγος ὁ αὐτός ἐστιν τῷ τῆς ΚΘ πρὸς τὴν ΕΞ, ὁ δὲ συνημμένος ἔκ τε τοῦ τῆς Γ Θ πρὸς τὴν ΘΚ καὶ τοῦ τῆς ΘΚ πρὸς τὴν ΕΞ μεταβαλλόμενος εἰς τὸν τῆς ΘΓ πρὸς ΕΞ λόγον, καὶ ὁ τῆς Γ Ζ πρὸς ΖΕ λόγος ὁ αὐτὸς τῷ τῆς ΓΘ πρὸς τὴν ΕΞ παραλλήλου οὔσης τῆς ΓΘ τῇ ΕΞ εὐθεῖα ἄρα ἐστὶν ἡ διὰ τῶν Θ, Ξ Ζ·
Since, as the product of ΑΖ, ΒΓ is to the product of ΑΒ, ΓΖ, so is the product of ΑΖ, ∠Ε to the product of Α∠, ΕΖ, by alternation, as the product of ΑΖ, ΒΓ is to the product of ΑΖ, ∠Ε, that is, as ΒΓ is to ∠Ε, so is the product of ΑΒ, ΓΖ to the product of Α∠, ΕΖ. But the ratio of ΒΓ to ∠Ε is compounded, if ΚΜ is drawn through Κ parallel to ΑΖ, of that of ΒΓ to ΚΝ and that of ΚΝ to ΚΜ and further of that of ΚΜ to ∠Ε, while the ratio of the product of ΑΒ, ΓΖ to the product of Α∠, ΕΖ is compounded of that of ΒΑ to Α∠ and that of ΓΖ to ΖΕ. Let the common ratio of ΒΑ to Α∠, which is the same as that of ΝΚ to ΚΜ, be cancelled; therefore, the remaining ratio of ΓΖ to ΖΕ is compounded of that of ΒΓ to ΚΝ, that is, of ΘΓ to ΚΘ, and that of ΚΜ to ∠Ε, that is, of ΚΗ to ΗΕ. Therefore, the (line) through Θ, Η, Ζ is a straight line. For if we draw ΕΞ through Ε parallel to Θ, and let the joined ΘΗ be produced to Ξ, the ratio of ΚΗ to ΗΕ is the same as that of ΚΘ to ΕΞ, and the ratio compounded of that of ΓΘ to ΘΚ and that of ΘΚ to ΕΞ is transformed into the ratio of ΘΓ to ΕΞ, and the ratio of ΓΖ to ΖΕ is the same as that of ΓΘ to ΕΞ, since ΓΘ is parallel to ΕΞ; therefore, the (line) through Θ, Ξ, Ζ is a straight line.
τοῦτο γὰρ φανερόν· ὥστε καὶ ἡ διὰ τῶν Θ, Η, Ζ εὐθεῖά ἐστιν.
For this is obvious; so that also the (line) through Θ, Η, Ζ is a straight line.
§6.5ε΄.
5.
Ἐὰν καταγραφὴ ἡ ΑΒΓ∠ΕΖΗΘ, γίνεται, ὡς ἡ Α∠ πρὸς τὴν ∠Γ, οὕτως ἡ ΑΒ πρὸς τὴν ΒΓ. ἔστω οὖν, ὡς ἡ Α∠ πρὸς τὴν ∠Γ. οὕτως ἡ Α πρὸς τὴν ΒΓ· ὅτι εὐθεῖά ἐστιν ἡ διὰ τῶν Α, Η, Θ.
ἤχθω διὰ τοῦ Η τῇ Α∠ παράλληλος ἡ ΚΛ. ἐπεὶ οὖν ἐστιν, ὡς ἡ Α∠ πρὸς τὴν ∠Γ, οὕτως ἡ ΑΒ πρὸς τὴν ΒΓ, ἀλλ᾿ ὡς μὲν ἡ Α∠ πρὸς τὴν ∠Γ οὕτως ἡ ΚΛ πρὸς τὴν ΛΗ, ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ ΚΗ πρὸς τὴν ΗΜ, καὶ ὡς ἄρα ἡ ΚΛ πρὸς τὴν ΛΗ, οὕτως ἡ ΚΗ πρὸς τὴν ΗΜ, καὶ λοιπὴ ἡ ΗΛ πρὸς λοιπὴν τὴν ΛΜ ἐστιν, ὡς ἡ ΚΛ πρὸς τὴν ΛΗ, τουτέστιν ὡς ἡ Α∠ πρὸς τὴν ∠Γ. ἐναλλάξ ἐστιν, ὡς ἡ Α∠ πρὸς τὴν ΗΛ, οὕτως ἡ Γ∠ πρὸς τὴν ΛΜ, τουτέστιν ἡ ∠Θ πρὸς ΘΛ. καί ἐστι παράλληλος ἡ ΗΛ τῇ Α∠· εὐθεῖα ἄρα ἐστὶν ἡ διὰ τῶν Α, Η, Θ σημείων·
If the figure is ΑΒΓ∠ΕΖΗΘ, as Α∠ is to ∠Γ, so ΑΒ becomes to ΒΓ. Let, therefore, as Α∠ is to ∠Γ, so be ΑΒ [text: Α] to ΒΓ; (to prove) that the (line) through Α, Η, Θ is a straight line. Let ΚΛ be drawn through Η parallel to Α∠. Since therefore, as Α∠ is to ∠Γ, so is ΑΒ to ΒΓ, but as Α∠ is to ∠Γ, so is ΚΛ to ΛΗ, and as ΑΒ is to ΒΓ, so is ΚΗ to ΗΜ, therefore also as ΚΛ is to ΛΗ, so is ΚΗ to ΗΜ, and the remaining ΗΛ is to the remaining ΛΜ as ΚΛ is to ΛΗ, that is, as Α∠ is to ∠Γ. By alternation, as Α∠ is to ΗΛ, so is Γ∠ to ΛΜ, that is, ∠Θ to ΘΛ. And ΗΛ is parallel to Α∠; therefore, the (line) through the points Α, Η, Θ is a straight line.
τοῦτο γὰρ φανερόν.
For this is obvious.