§6.10ι΄.
10.
Εἰς δύο εὐθείας τὰς ΒΑΕ, ∠ΑΗ ἀπὸ τοῦ Θ σημείου δύο διήχθωσαν εὐθεῖαι αἱ ∠Θ, ΘΕ, ἔστω δέ, ὡς τὸ ὑπὸ τῶν ∠Θ, ΒΓ πρὸς τὸ ὑπὸ ∠Γ, ΒΘ, οὕτως τὸ ὑπὸ ΘΗ, ΖΕ πρὸς τὸ ὑπὸ ΘΕ, ΖΗ ὅτι εὐθεῖά ἔστιν ἡ διὰ τῶν Α, Ζ.
ἤχθω διὰ τοῦ Θ τῇ ΓΑ παράλληλος ἡ ΚΛ καὶ συμπιπτέτω ταῖς ΑΒ, Α∠ κατὰ τὰ Κ, Λ σημεῖα, καὶ διὰ τοῦ Λ τῇ Α∠ παράλληλος ἤχθω ἡ ΛΜ, καὶ ἐκβεβλήσθω ἡ ΕΘ ἐπὶ τὸ Μ, διὰ δὲ τοῦ Κ τῇ ΑΒ παράλληλος ἤχθω ἡ ΚΝ, καὶ ἐκβεβλήσθω ἡ ∠Θ ἐπὶ τὸ Ν.
ἐπεὶ οὖν διὰ τὰς παραλλήλους γίνεται, ὡς ἡ ∠Θ πρὸς τὴν ΘΝ, οὕτως ἡ ∠Γ πρὸς τὴν ΓΒ, τὸ ἄρα ὑπὸ τῶν ∠Θ, ΓΒ ἴσον ἐστὶν τῷ ὑπὸ τῶν ∠Γ, ΘΝ. ἄλλο δέ τι τυχὸν τὸ ὑπὸ ∠Γ ΒΘ· ἔστιν ἄρα, ὡς τὸ ὑπὸ ∠Θ, ΒΓ πρὸς τὸ ὑπὸ ∠Γ ΒΘ, οὕτως τὸ ὑπὸ Γ∠, ΘΝ πρὸς τὸ ὑπὸ ∠Γ ΒΘ, τουτέστιν ἡ ΘΝ πρὸς ΘΒ. ἀλλ᾿ ὡς μὲν τὸ ὑπὸ Θ∠, ΒΓ πρὸς τὸ ὑπὸ ∠Γ. ΒΘ, ὑπόκειται τὸ ὑπὸ ΘΗ, ΖΕ πρὸς τὸ ὑπὸ ΘΕ, ΖΗ, ὡς δὲ ἡ ΘΝ πρὸς ΘΒ, οὕτως ἡ ΚΘ πρὸς ΘΛ, τουτέστιν ἐν παραλλήλῳ ἡ ΗΘ πρὸς τὴν ΘΜ, τουτέστιν τὸ ὑπὸ ΘΗ, ΖΕ πρὸς τὸ ὑπὸ ΘΜ, ΖΕ καὶ ὡς ἄρα τὸ ὑπὸ ΘΗ, ΖΕ πρὸς τὸ ὑπὸ ΘΕ, ΖΗ, οὕτως ἐστὶν τὸ ὑπὸ ΘΗ, ΖΕ πρὸς τὸ ὑπὸ ΘΜ, ΖΕ ἴσον ἄρα ἐστὶν τὸ ὑπὸ ΘΕ, ΖΗ τῷ ὑπὸ ΘΜ, ΖΕ. καὶ ἐπεί ἐστιν, ὡς ἡ ΘΜ πρὸς τὴν ΘΕ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΕ, συνθέντι καὶ ἐναλλάξ ἐστιν, ὡς ἡ ΜΕ πρὸς τὴν ΕΗ, οὕτως ἡ ΘΕ πρὸς τὴν ΕΖ. ἀλλʼ ὡς ἡ ΜΕ πρὸς τὴν ΕΗ, οὕτως ἐστὶν ἡ ΛΕ πρὸς τὴν ΕΑ· καὶ ὡς ἄρα ἡ ΛΕ πρὸς τὴν ΕΑ, οὕτως ἡ ΘΕ πρὸς τὴν ΕΖ·
To two straight lines ΒΑΕ, ∠ΑΗ, let two straight lines ∠Θ, ΘΕ be drawn from the point Θ, and let, as the rectangle contained by ∠Θ, ΒΓ is to the rectangle contained by ∠Γ, ΒΘ, so the rectangle contained by ΘΗ, ΖΕ is to the rectangle contained by ΘΕ, ΖΗ; (to prove) that the line through Α, Ζ is a straight line. Let ΚΛ be drawn through Θ parallel to ΓΑ, and let it meet ΑΒ, Α∠ at the points Κ, Λ, and through Λ let ΛΜ be drawn parallel to Α∠, and let ΕΘ be produced to Μ, and through Κ let ΚΝ be drawn parallel to ΑΒ, and let ∠Θ be produced to Ν. Since therefore, because of the parallels, as ∠Θ is to ΘΝ, so is ∠Γ to ΓΒ, therefore the rectangle contained by ∠Θ, ΓΒ is equal to the rectangle contained by ∠Γ, ΘΝ. And let the rectangle contained by ∠Γ, ΒΘ be some other arbitrary magnitude; therefore, as the rectangle ∠Θ, ΒΓ is to the rectangle ∠Γ, ΒΘ, so is the rectangle Γ∠, ΘΝ to the rectangle ∠Γ, ΒΘ, that is, ΘΝ to ΘΒ. But as the rectangle Θ∠, ΒΓ is to the rectangle ∠Γ, ΒΘ, there is assumed the rectangle ΘΗ, ΖΕ to the rectangle ΘΕ, ΖΗ, and as ΘΝ is to ΘΒ, so is ΚΘ to ΘΛ, that is, between parallels, ΗΘ to ΘΜ, that is, the rectangle ΘΗ, ΖΕ to the rectangle ΘΜ, ΖΕ; and therefore, as the rectangle ΘΗ, ΖΕ is to the rectangle ΘΕ, ΖΗ, so is the rectangle ΘΗ, ΖΕ to the rectangle ΘΜ, ΖΕ; therefore the rectangle ΘΕ, ΖΗ is equal to the rectangle ΘΜ, ΖΕ. And since, as ΘΜ is to ΘΕ, so is ΗΖ to ΖΕ, by composition and alternation, as ΜΕ is to ΕΗ, so is ΘΕ to ΕΖ. But as ΜΕ is to ΕΗ, so is ΛΕ to ΕΑ; and therefore as ΛΕ is to ΕΑ, so is ΘΕ to ΕΖ; therefore ΑΖ is parallel to ΚΛ.
παράλληλος ἄρα ἐστὶν ἡ ΑΖ τῇ ΚΛ. ἀλλὰ καὶ ἡ ΓΑ εὐθεῖα ἄρα ἐστὶν ἡ ΓΑΖ·
But ΚΛ is also parallel to ΓΑ; therefore ΓΑΖ is a straight line; which was to be proved.
ὅπερ ἔδει δεῖξαι. τὰ δὲ πτωτικὰ αὐτοῦ ὁμοίως τοῖς προγεγραμμένοις, ὧν ἐστιν ἀναστρόφιον.
And its cases are similar to those written before, of which they are the converse.
§6.11ια΄.
11.
Τρίγωνον τὸ ΑΒΓ. καὶ τῇ ΒΓ παράλληλος ἡ Α∠, καὶ διαχθεῖσα ἡ ∠Ε τῇ ΒΓ συμπιπτέτω κατὰ τὸ Ε σημεῖον· ὅτι ἐστίν, ὡς τὸ ὑπὸ ∠Ε, ΖΗ πρὸς τὸ ὑπὸ ΕΖ, Η∠, οὕτως ἡ ΓΒ πρὸς τὴν ΒΕ.
ἤχθω διὰ τοῦ Γ τῇ ∠Ε παράλληλος ἡ ΓΘ, καὶ ἐκβεβλήσθω ἡ ΑΒ ἐπὶ τὸ Θ.
ἐπεὶ οὖν ἐστιν, ὡς ἡ Γ πρὸς τὴν ΑΗ, οὕτως ἡ ΓΘ πρὸς τὴν ΖΗ, ὡς δὲ ἠ Γ πρὸς τὴν ΑΗ, οὕτως ἐστὶν ἡ Ε∠ πρὸς τὴν ∠Η. καὶ ὡς ἄρα ἡ Ε∠ πρὸς τὴν ∠Η, οὕτως ἐστὶν ἡ ΘΓ πρὸς τὴν ΖΗ τὸ ἄρα ὑπὸ τῶν ΓΘ, ∠Η ἴσον ἐστὶν τῷ ὑπὸ τῶν Ε∠, ΖΗ. ἄλλο δέ τι τυχὸν τὸ ὑπὸ ΕΖ, Η∠ ἔστιν ἄρα, ὡς τὸ ὑπὸ ∠Ε, ΖΗ πρὸς τὸ ὑπὸ ∠Η, ΕΖ, οὕτως τὸ ὑπὸ ΓΘ, ∠Η πρὸς τὸ ὑπὸ ∠Η, ΕΖ, τουτέστιν ἡ ΓΘ πρὸς ΕΖ, τσυτέστιν ἡ ΓΒ πρὸς ΒΕ. ἔστιν οὖν, ὡς τὸ ὑπὸ ∠Ε, ΖΗ πρὸς τὸ ὑπὸ ΕΖ, Η∠, οὕτως ἡ ΓΒ πρὸς ΒΕ.
τὰ δʼ αὐτά, κἂν ἐπὶ τὰ ἕτερα μέρη ἀχθῇ ἡ Α∠ παράλληλος, καὶ ἀπὸ τοῦ ∠ ἐκτὸς ὡς ἐπὶ τὸ Γ διαχθῇ ἡ εὐθεῖα.
Let there be a triangle ΑΒΓ. And let Α∠ be parallel to ΒΓ, and let the straight line ∠Ε drawn from ∠ meet ΒΓ at the point Ε; (to prove) that, as the rectangle ∠Ε, ΖΗ is to the rectangle ΕΖ, Η∠, so is ΓΒ to ΒΕ. Let ΓΘ be drawn through Γ parallel to ∠Ε, and let ΑΒ be produced to Θ. Since therefore, as Γ is to ΑΗ, so is ΓΘ to ΖΗ, and as Γ is to ΑΗ, so is Ε∠ to ∠Η, therefore also as Ε∠ is to ∠Η, so is ΘΓ to ΖΗ; therefore the rectangle ΓΘ, ∠Η is equal to the rectangle Ε∠, ΖΗ. And let the rectangle ΕΖ, Η∠ be some other arbitrary magnitude; therefore, as the rectangle ∠Ε, ΖΗ is to the rectangle ∠Η, ΕΖ, so is the rectangle ΓΘ, ∠Η to the rectangle ∠Η, ΕΖ, that is, ΓΘ to ΕΖ, that is, ΓΒ to ΒΕ. Therefore, as the rectangle ∠Ε, ΖΗ is to the rectangle ΕΖ, Η∠, so is ΓΒ to ΒΕ. And the same holds even if Α∠ is drawn parallel to the other parts, and the straight line is drawn from ∠ outwards as if to Γ.