§6.15ιεʹ.
15.
Τούτου προτεθεωρημένου ἔστω παράλληλος ἡ ΑΒ τῇ Γ∠, καὶ εἰς αὐτὰς ἐμπιπτέτωσαν εὐθεῖαι αἱ ΑΖ, ΖΒ, ΓΕ, Ε∠, καὶ ἐπεζεύχθωσαν αἱ ΒΓ, ΗΚ ὅτι εὐθεῖά ἐστιν ἡ διὰ τῶν Α, Μ, ∠.
This having been preliminarily proved, let ΑΒ be parallel to Γ∠, and let straight lines ΑΖ, ΖΒ, ΓΕ, Ε∠ fall upon them, and let ΒΓ, ΗΚ be joined; (to prove) that the line through Α, Μ, ∠ is a straight line.
Ἐπεζεύχθω ἡ ∠Μ καὶ ἐκβεβλήσθω ἐπὶ τὸ Θ.
ἐπεὶ οὖν τριγώνου τοῦ ΒΓΖ ἐκτὸς ἀπὸ τῆς κορυφῆς τοῦ Β σημείου τῇ Γ∠ παράλληλος ἦκται ἡ ΒΕ, καὶ διῆκται ἡ ∠ Ε, γίνεται, ὡς ἡ Γ Ζ πρὸς Ζ∠, οὕτως τὸ ὑπὸ ∠Ε, ΚΛ πρὸς τὸ ὑπὸ ΕΛ, Κ∠.
Let ∠Μ be joined and produced to Θ. Since therefore, outside the triangle ΒΓΖ, from the vertex point Β, ΒΕ has been drawn parallel to Γ∠, and ∠Ε has been drawn through, it follows that, as ΓΖ is to Ζ∠, so is the rectangle contained by ∠Ε, ΚΛ to the rectangle contained by ΕΛ, Κ∠.
ὡς δὲ τὸ ὑπὸ ∠Ε, ΚΛ πρὸς τὸ ὑπὸ ∠ Κ, ΛΕ, οὕτως ἐστὶν τὸ ὑπὸ ΓΗ, ΘΕ πρὸς τὸ ὑπὸ ΓΕ, ΗΘ, ἐπεὶ εἰς τρεῖς εὐθείας τὰς ΓΛ, ∠Θ, ΗΚ δύο εἰσὶν διηγμέναι ἀπὸ τοῦ αὐτοῦ σημείου τοῦ Ε α ΕΓ, Ε∠ καὶ ὡς ἄρα ἡ ∠Ζ πρὸς ΖΓ οὕτως ἐστὶν τὸ ὑπὸ ΓΕ, ΗΘ πρὸς τὸ ὑπὸ ΓΗ ΘΕ. διὰ τὸ προγεγραμμένον ἄρα ἡ διὰ τῶν Α, Μ, ∠ ἐστιν εὐθεῖα.
And as the rectangle contained by ∠Ε, ΚΛ is to the rectangle contained by ∠Κ, ΛΕ, so is the rectangle contained by ΓΗ, ΘΕ to the rectangle contained by ΓΕ, ΗΘ, since to three straight lines ΓΛ, ∠Θ, ΗΚ, two straight lines ΕΓ, Ε∠ have been drawn from the same point Ε. Therefore also, as ∠Ζ is to ΖΓ, so is the rectangle contained by ΓΕ, ΗΘ to the rectangle contained by ΓΗ, ΘΕ. Therefore, by what was written before, the line through Α, Μ, ∠ is a straight line.
§6.16ιϚ΄.
16.
Εἰς δύο εὐθείας τὰς ΑΒ, ΑΓ ἀπὸ τοῦ αὐτοῦ σημείου τοῦ ∠ δύο διήχθωσαν αἱ ∠Β, ∠Ε, καὶ ἐπʼ αὐτῶν εἰλήφθω σημεῖα τὰ Η, Θ, ἔστω δέ, ὡς τὸ ὑπὸ ΕΗ, Ζ∠ πρὸς τὸ ὑπὸ ∠Ε, ΗΖ, οὕτως τὸ ὑπὸ ΒΘ, Γ∠ πρὸς τὸ ὑπὸ Β∠, ΓΘ ὅτι εὐθεῖά ἐστιν ἡ διὰ τῶν Α, Η, Θ.
ἤχθω διὰ τοῦ Η τῇ Β∠ παράλληλος ἡ ΚΛ.
ἐπεὶ οὖν ἐστιν, ὡς τὸ ὑπὸ ΕΗ, Ζ∠ πρὸς τὸ ὑπὸ ∠Ε, ΖΗ, οὕτως τὸ ὑπὸ ΒΘ, Γ∠ πρὸς τὸ ὑπὸ Β∠, ΓΘ, ἀλλὰ ὁ τοῦ ὑπὸ ΕΗ, Ζ∠ πρὸς τὸ ὑπὸ ∠Ε, ΗΖ συνῆπται λόγος ἔκ τε τοῦ ὃν ἔχει ἡ ΗΕ πρὸς Ε∠, τουτέστιν ἡ ΚΗ πρὸς Β∠, καὶ ἐξ οὗ ὃν ἔχει ἡ ∠Ζ πρὸς ΖΗ, τουτέστιν ἡ Γ∠ πρὸς τὴν ΗΛ, ὁ δὲ τοῦ ὑπὸ ΒΘ, Γ∠ πρὸς τὸ ὑπὸ Β∠, ΓΘ συνῆπται λόγος ἔκ τε τοῦ ὃν ἔχει ἡ ΘΒ πρὸς Β∠ καὶ ἐξ οὗ ὃν ἔχει ἡ ∠Γ πρὸς ΓΘ, καὶ ὁ ἔκ τε τοῦ τῆς Κ ἄρα πρὸς Β∠ καὶ τοῦ τῆς ∠Γ πρὸς ΗΛ ὁ αὐτός ἐστιν τῷ συνημμένῳ ἔκ τε τοῦ τῆς ΒΘ πρὸς Β∠ καὶ τοῦ τῆς ∠Γ πρὸς ΓΘ. ὁ δὲ τῆς ΚΗ πρὸς Β∠ συνῆπται ἔκ τε τοῦ τῆς ΚΗ πρὸς ΒΘ καὶ τοῦ τῆς ΒΘ πρὸς Β∠· ὁ ἄρα συνημμένος ἔκ τε τοῦ τῆς ΚΗ πρὸς ΒΘ καὶ τοῦ τῆς ΒΘ πρὸς Β∠ καὶ ἔτι τοῦ τῆς ∠Γ πρὸς ΗΛ ὁ αὐτός ἐστιν τῷ συνημμένῳ ἔκ τε τοῦ τῆς ΒΘ πρὸς Β∠ καὶ τοῦ τῆς ∠Γ πρὸς ΓΘ. κοινὸς ἐκκεκρούσθω ὁ τῆς Θ Β πρὸς Β∠ λόγος·
To two straight lines ΑΒ, ΑΓ, from the same point ∠, let two straight lines ∠Β, ∠Ε be drawn, and let points Η, Θ be taken on them, and let it be that, as the rectangle contained by ΕΗ, Ζ∠ is to the rectangle contained by ∠Ε, ΗΖ, so is the rectangle contained by ΒΘ, Γ∠ to the rectangle contained by Β∠, ΓΘ; (to prove) that the line through Α, Η, Θ is a straight line. Let ΚΛ be drawn through Η parallel to Β∠. Since therefore, as the rectangle contained by ΕΗ, Ζ∠ is to the rectangle contained by ∠Ε, ΖΗ, so is the rectangle contained by ΒΘ, Γ∠ to the rectangle contained by Β∠, ΓΘ, but the ratio of the rectangle contained by ΕΗ, Ζ∠ to the rectangle contained by ∠Ε, ΗΖ is compounded of that which ΗΕ has to Ε∠, that is, ΚΗ to Β∠, and of that which ∠Ζ has to ΖΗ, that is, Γ∠ to ΗΛ, and the ratio of the rectangle contained by ΒΘ, Γ∠ to the rectangle contained by Β∠, ΓΘ is compounded of that which ΘΒ has to Β∠ and of that which ∠Γ has to ΓΘ, therefore the ratio compounded of that of ΚΗ to Β∠ and that of ∠Γ to ΗΛ is the same as the ratio compounded of that of ΒΘ to Β∠ and that of ∠Γ to ΓΘ. But the ratio of ΚΗ to Β∠ is compounded of that of ΚΗ to ΒΘ and that of ΒΘ to Β∠; therefore the ratio compounded of that of ΚΗ to ΒΘ and that of ΒΘ to Β∠ and further that of ∠Γ to ΗΛ is the same as the ratio compounded of that of ΒΘ to Β∠ and that of ∠Γ to ΓΘ.
λοιπὸς ἄρα ὁ συνημμένος ἔκ τε τοῦ τῆς ΚΗ πρὸς ΒΘ καὶ τοῦ τῆς ∠Γ πρὸς ΚΛ ὁ αὐτός ἐστιν τῷ τῆς ∠Γ πρὸς τὴν ΓΘ, τουτέστιν τῷ συνημμένῳ ἔκ τε τοῦ τῆς ∠Γ πρὸς τὴν ΗΛ καὶ τοῦ τῆς ΗΛ πρὸς τὴν ΘΓ. καὶ πάλιν κοινὸς ἐκκεκρούσθω ὁ τῆς ∠Γ πρὸς τὴν ΗΛ λόγος·
Let the common ratio of ΘΒ to Β∠ be struck out; therefore, the remaining ratio compounded of that of ΚΗ to ΒΘ and that of ∠Γ to ΗΛ is the same as the ratio of ∠Γ to ΓΘ, that is, the ratio compounded of that of ∠Γ to ΗΛ and that of ΗΛ to ΘΓ. And again, let the common ratio of ∠Γ to ΗΛ be struck out; therefore, the remaining ratio of ΚΗ to ΒΘ is the same as the ratio of ΗΛ to ΘΓ.
λοιπὸς ἄρα ὁ τῆς ΚΗ πρὸς τὴν ΒΘ λόγος ὁ αὐτός ἐστιν τῷ τῆς ΗΛ πρὸς τὴν ΘΓ. καὶ ἐναλλάξ ἐστιν, ὡς ἡ ΚΗ πρὸς τὴν ΗΛ, οὕτως ἡ ΒΘ πρὸς τὴν ΘΓ. καί εἴσιν αἱ ΚΛ, ΒΓ παράλληλοι· εὐθεῖα ἄρα ἐστὶν ἡ διὰ τῶν Α, Η, Θ σημείων.
And alternately, as ΚΗ is to ΗΛ, so is ΒΘ to ΘΓ. And ΚΛ, ΒΓ are parallel; therefore the line through the points Α, Η, Θ is a straight line.