§6.6ϛ΄.
6.
Πάλιν, ἐὰν καταγραφή, καὶ παράλληλος ἡ ∠Ζ τῇ ΒΓ, γίνεται ἴση ἡ ΑΒ τῇ ΒΓ ἔστω οὖν ἴση·
Again, if (there is) a figure, and ∠Ζ is parallel to ΒΓ, ΑΒ becomes equal to ΒΓ.
ὅτι παράλληλος. ἔστιν δέ·
Let it therefore be equal; (to prove) that (it is) parallel.
ἐὰν γὰρ τῇ ΕΒ προσθῶ τῇ Ηβ ἴσην τὴν ΒΘ καὶ ἐπιζεύξω τὰς ΑΘ, ΘΓ, γίνεται παραλληλόγραμμον τὸ ΑΘΓΗ, καὶ διὰ τοῦτό ἐστιν, ὡς ἡ Α∠ πρὸς τὴν ∠Ε, οὕτως ἡ Γ πρὸς τὴν ΖΕ· ἑκάτερος γὰρ τῶν εἰρημένων ὁ αὐτός ἐστιν τῷ τῆς Θ πρὸς τὴν ΗΕ λόγῳ· ὥστε παράλληλός ἐστιν ἡ ∠ τῇ ΑΓ.
And it is (as follows): for if to ΕΒ I add ΒΘ equal to ΗΒ [text: Ηβ], and join ΑΘ, ΘΓ, ΑΘΓΗ becomes a parallelogram, and because of this, as Α∠ is to ∠Ε, so is Γ to ΖΕ; for each of the said (ratios) is the same as the ratio of Θ to ΗΕ; so that ∠ is parallel to ΑΓ.
§6.7ζ΄.
7.
Ἔστω καταγραφή, καὶ τῶν ∠Β, ΒΓ μέση ἀνάλογον ἔστω ἡ ΒΑ ὅτι παράλληλός ἔστιν ἡ ΖΗ τῇ ΑΓ.
ἐκβεβλήσθω ἡ ΕΒ, καὶ διὰ τοῦ Α τῇ ∠Ζ εὐθείᾳ παράλληλος ἤχθω ἡ ΑΚ, καὶ ἐπεζεύχθω ἡ ΓΚ.
ἐπεὶ οὖν ἔστιν, ὡς ἡ Γ Β πρὸς τὴν ΒΑ, οὕτως ἡ ΑΒ πρὸς τὴν Β∠, ὡς δὲ ἡ ΑΒ πρὸς τὴν Β∠, οὕτως ἡ ΚΒ πρὸς τὴν ΒΘ, καὶ ὡς ἄρα ἡ ΓΒ πρὸς τὴν ΒΑ, οὕτως ἡ Κ Β πρὸς τὴν ΒΘ·
Let there be a figure, and let ΒΑ be a mean proportional between ∠Β, ΒΓ; (to prove) that ΖΗ is parallel to ΑΓ. Let ΕΒ be produced, and through Α let ΑΚ be drawn parallel to the straight line ∠Ζ, and let ΓΚ be joined. Since therefore, as ΓΒ is to ΒΑ, so is ΑΒ to Β∠, and as ΑΒ is to Β∠, so is ΚΒ to ΒΘ, and therefore as ΓΒ is to ΒΑ, so is ΚΒ to ΒΘ; therefore ΑΘ is parallel to ΚΓ.
παράλληλος ἄρα ἐστὶν ἡ ΑΘ τῇ ΚΓ ἔστιν οὖν πάλιν, ὡς ἡ Α πρὸς τὴν ΖΕ, οὕτως ἡ ΓΗ πρὸς τὴν ΗΕ· ἑκάτερος γὰρ τῶν εἰρημένων λόγος ὁ αὐτός ἐστιν τῷ τῆς ΚΘ πρὸς τὴν ΘΕ ὥστε παράλληλός ἐστιν ἡ ΖΗ τῇ Α∠.
Therefore again, as Α is to ΖΕ, so is ΓΗ to ΗΕ; for each of the said ratios is the same as that of ΚΘ to ΘΕ; so that ΖΗ is parallel to Α∠.
§6.8η΄.
8.
Ἔστω βωμίσκος ὁ ΑΒΓ∠ΕΖΗ, καὶ ἔστω παράλληλος ἡ μὲν ∠Ε τῇ ΒΓ, ἡ δὲ ΕΗ τῇ ΒΖ ὅτι καὶ ἡ ∠Ζ τῇ ΓΗ παράλληλός ἐστιν.
Let there be a small altar ΑΒΓ∠ΕΖΗ, and let ∠Ε be parallel to ΒΓ, and ΕΗ to ΒΖ; (to prove) that ∠Ζ is also parallel to ΓΗ.
ἐπεζεύχθωσαν αἱ ΒΕ, ∠Γ, ΖΗ ἴσον ἄρα ἐστὶν τὸ ∠ΒΕ τρίγωνον τῷ ∠ΓΕ τριγώνῳ.
Let ΒΕ, ∠Γ, ΖΗ be joined; therefore triangle ∠ΒΕ is equal to triangle ∠ΓΕ.
κοινὸν προσκείσθω τὸ ∠ΑΕ τρίγωνον· ὅλον ἄρα τὸ ΑΒΕ τρίγωνον ὅλῳ τῷ
Γ∠Α τριγώνῳ ἴσον ἐστίν.
Let the common triangle ∠ΑΕ be added; therefore the whole triangle ΑΒΕ is equal to the whole triangle Γ∠Α.
πάλιν, ἐπεὶ παράλληλός ἔστιν ἡ ΒΖ τῇ ΕΗ, ἴσον ἐστὶν τὸ ΒΖΕ τρίγωνον τῷ ΒΖΗ τριγώνῳ.
Again, since ΒΖ is parallel to ΕΗ, triangle ΒΖΕ is equal to triangle ΒΖΗ.
κοινὸν ἀφῃρήσθω τὸ ΑΒ Ζ τρίγωνον· λοιπὸν ἄρα τὸ ΑΒΕ τρίγωνον λοιπῷ τῷ ΑΗΖ τριγώνῳ ἴσον ἐστίν.
Let the common triangle ΑΒΖ be subtracted; therefore the remaining triangle ΑΒΕ is equal to the remaining triangle ΑΗΖ.
ἀλλὰ τὸ ΑΒΕ τρίγωνον τῷ ΑΓ∠ τριγώνῳ ἐστὶν ἴσον· καὶ τὸ ΑΓ∠ ἄρα τρίγωνον τῷ ΑΖΗ τριγώνῳ ἴσον ἐστίν.
But triangle ΑΒΕ is equal to triangle ΑΓ∠; therefore triangle ΑΓ∠ is also equal to triangle ΑΖΗ.
κοινὸν προσκείσθω τὸ ΑΓΗ τρίγωνον· ὅλον ἄρα τὸ Γ∠Η τρίγωνον ὅλῳ τῷ ΓΖΗ τριγώνῳ ἴσον ἐστίν.
Let the common triangle ΑΓΗ be added; therefore the whole triangle Γ∠Η is equal to the whole triangle ΓΖΗ.
καί ἐστιν ἐπὶ τῆς αὐτῆς βάσεως τῆς ΓΗ· παράλληλος ἄρα ἐστὶν ἡ ΓΗ τῇ ∠Ζ.
And they are on the same base ΓΗ; therefore ΓΗ is parallel to ∠Ζ.
§6.9θ΄.
9.
Ἔστω τρίγωνον τὸ ΑΒΓ καὶ ἐν αὐτῷ διήχθωσαν αἱ Α∠, ΑΕ, καὶ τῇ ΒΓ παράλληλος ἤχθω ἡ ΖΗ, καὶ κεκλάσθω ἡ ΖΘΗ, ἔστω δέ, ὡς ἡ ΒΘ πρὸς τὴν ΘΓ, οὕτως ἡ ∠Θ πρὸς τὴν ΘΕ ὅτι παράλληλός ἔστιν ἡ ΚΛ τῇ ΒΓ.
ἐπεὶ γάρ ἐστιν, ὡς ἡ ΒΘ πρὸς τὴν ΘΓ, οὕτως ἡ ∠Θ πρὸς τὴν ΘΕ, λοιπὴ ἄρα ἡ Β∠ πρὸς λοιπὴν τὴν ΓΕ ἐστιν, ὡς ἡ ∠Θ πρὸς τὴν ΘΕ. ὡς δὲ ἡ Β∠ πρὸς τὴν ΕΓ, οὕτως ἐστὶν ἡ ΖΜ πρὸς Ν καὶ ὡς ἄρα ἡ ΖΜ πρὸς ΝΗ, οὕτως ἐστὶν ἡ ∠Θ πρὸς τὴν ΘΕ. ἐναλλάξ ἔστιν, ὡς ἡ ΖΜ πρὸς τὴν ∠Θ, οὕτως ἡ ΝΗ πρὸς τὴν ΘΕ. ἀλλʼ ὡς μὲν ἡ ΖΜ πρὸς τὴν ∠Θ, οὕτως ἐστὶν ἐν παραλλήλῳ ἡ ΖΚ πρὸς τὴν ΚΘ, ὡς δὲ ἡ ΗΝ πρὸς τὴν ΘΕ, οὕτως ἐστὶν ἡ ΗΛ πρὸς τὴν ∠Θ καὶ ὡς ἄρα ἡ ΖΚ πρὸς τὴν Κ Θ, οὕτως ἐστὶν ἡ ΗΛ πρὸς τὴν ∠Θ. παράλληλος ἄρα ἐστὶν ἡ ΚΛ τῇ ΗΖ· ὥστε καὶ τῇ ΓΒ.
Let there be a triangle ΑΒΓ, and in it let Α∠, ΑΕ be drawn, and let ΖΗ be drawn parallel to ΒΓ, and let the broken line ΖΘΗ be drawn, and let, as ΒΘ is to ΘΓ, so ∠Θ is to ΘΕ; (to prove) that ΚΛ is parallel to ΒΓ. For since, as ΒΘ is to ΘΓ, so is ∠Θ to ΘΕ, therefore the remaining Β∠ is to the remaining ΓΕ as ∠Θ is to ΘΕ. And as Β∠ is to ΕΓ, so is ΖΜ to Ν, and therefore as ΖΜ is to ΝΗ, so is ∠Θ to ΘΕ. By alternation, as ΖΜ is to ∠Θ, so is ΝΗ to ΘΕ. But as ΖΜ is to ∠Θ, so, between parallels, is ΖΚ to ΚΘ, and as ΗΝ is to ΘΕ, so is ΗΛ to ∠Θ; and therefore as ΖΚ is to ΚΘ, so is ΗΛ to ∠Θ. Therefore, ΚΛ is parallel to ΗΖ; so that also to ΓΒ.