§6.23κγ΄.
23.
Ἔστω τὸ ὑπὸ ΑΒΓ ἴσον τῷ ἀπὸ Β∠ τετραγώνῳ· ὅτι γίνεται γ, τὸ μὲν ὑπὸ συναμφοτέρου τῆς Α∠Γ καὶ τῆς Β∠ ἴσον τῷ ὑπὸ Α∠, ∠Γ, τὸ δὲ ὑπὸ συναμφοτέρου τῆς Α∠Γ καὶ τῆς ΒΓ ἴσον τῷ ἀπὸ ∠Γ τετραγώνῳ, τὸ δὲ ὑπὸ συναμφοτέρου τῆς Α∠Γ καὶ τῆς ΒΑ ἴσον τῷ ἀπὸ Α∠ τετραγώνῳ.
Let the rectangle contained by ΑΒ, ΒΓ be equal to the square on Β∠; (to prove) that three things follow: first, the rectangle contained by the sum of Α∠, ∠Γ and Β∠ is equal to the rectangle contained by Α∠, ∠Γ; second, the rectangle contained by the sum of Α∠, ∠Γ and ΒΓ is equal to the square on ∠Γ; third, the rectangle contained by the sum of Α∠, ∠Γ and ΒΑ is equal to the square on Α∠.
ἐπεὶ γὰρ τὸ ὑπὸ ΑΒΓ ἴσον ἐστὶν τῷ ἀπὸ Β∠, ἀνάλογον καὶ ὅλη πρὸς ὅλην καὶ ἀνάπαλιν καὶ συνθέντι· ἔστιν ἄρα, ὡς συναμφότερος ἡ Γ∠, ∠Α πρὸς τὴν ∠Α, οὕτως ἡ Γ∠ πρὸς τὴν ∠Β·
For since the rectangle contained by ΑΒ, ΒΓ is equal to the square on Β∠, whole is proportional to whole, and by inversion and addition; therefore, as the sum of Γ∠, ∠Α is to ∠Α, so is Γ∠ to ∠Β; therefore, the rectangle contained by the sum of Α∠, ∠Γ and Β∠ is equal to the rectangle contained by Α∠, ∠Γ.
τὸ ἄρα ὑπὸ συναμφοτέρου τῆς Α∠, ∠Γ καὶ τῆς Β∠ ἴσον ἐστὶ τῷ ὑπὸ τῶν Α∠Γ. πάλιν, ἐπεὶ ὅλη ἡ Α∠ πρὸς ὅλην τὴν ∠Γ ἐστιν, ὡς ἡ ∠Β πρὸς τὴν ΒΓ, συνθέντι ἐστίν, ὡς συναμφότερος ἡ Α∠Γ πρὸς τὴν ∠Γ, οὕτως ἡ ∠Γ πρὸς τὴν ΓΒ·
Again, since the whole Α∠ is to the whole ∠Γ, as ∠Β is to ΒΓ, by addition, as the sum of Α∠, ∠Γ is to ∠Γ, so is ∠Γ to ΓΒ; therefore, the rectangle contained by the sum of Α∠, ∠Γ and ΓΒ is equal to the square on ∠Γ.
τὸ ἄρα ὑπὸ συναμφοτέρου τῆς Α∠Γ καὶ τῆς ΓΒ ἴσον ἐστὶν τῷ ἀπὸ ∠Γ. πάλιν, ἐπεὶ ὅλη ἡ Α∠ πρὸς ὅλην τὴν ∠Γ ἐστιν, ὡς ἡ ΑΒ πρὸς τὴν Β∠, ἀνάπαλιν καὶ συνθέντι ἐστίν, ὡς συναμφότερος ἡ Γ∠Α πρὸς τὴν ∠Α, οὕτως ἡ ∠Α πρὸς τὴν ΑΒ· τὸ ἄρα ὑπὸ συναμφοτέρου τῆς Α∠Γ καὶ τῆς ΑΒ ἴσον ἐστὶν τῷ ἀπὸ Α∠ τετραγώνῳ.
Again, since the whole Α∠ is to the whole ∠Γ, as ΑΒ is to Β∠, by inversion and addition, as the sum of Γ∠, ∠Α is to ∠Α, so is ∠Α to ΑΒ; therefore, the rectangle contained by the sum of Α∠, ∠Γ and ΑΒ is equal to the square on Α∠.
§6.24κδ΄.
24.
Εὐθεῖα ἡ ΑΒ καὶ δύο σημεῖα τὰ Γ, Δ, καὶ ἔστω τὸ ἀπὸ Γ∠ τετράγωνον ἴσον τῷ δὲς ὑπὸ ΑΓ Β∠· ὅτι καὶ τὸ ἀπὸ ΑΒ τετράγωνον ἴσον ἐστὶν τοῖς ἀπὸ τῶν Α∠, ΓΒ τετραγώνοις.
Let there be a straight line ΑΒ and two points Γ, Δ, and let the square on Γ∠ be equal to twice the rectangle contained by ΑΓ, Β∠; (to prove) that the square on ΑΒ is also equal to the squares on Α∠, ΓΒ.
ἐπεὶ γὰρ τὸ ἀπὸ Γ∠ ἴσον ἐστὶν τῷ θὶς ὑπὸ ΑΓ, ∠Β, τὸ ἄρα δίς ὑπὸ ΑΓΒ ἴσον ἐστὶν τῷ τε ἀπὸ τῆς Γ∠ καὶ τῷ δὶς ὑπὸ τῶν ΑΓ∠.
For since the square on Γ∠ is equal to twice the rectangle contained by ΑΓ, ∠Β, therefore twice the rectangle contained by ΑΓ, ΓΒ is equal to both the square on Γ∠ and twice the rectangle contained by ΑΓ, ∠.
κοινὸν προσκείσθω τὸ ἀπὸ ΑΓ· τὸ ἄρα δὶς ὑπὸ ΑΓΒ μετὰ τοῦ ἀπὸ ΑΓ ἴσον ἐστὶν τῷ ἀπὸ Α∠.
Let the common square on ΑΓ be added; therefore twice the rectangle contained by ΑΓ, ΓΒ together with the square on ΑΓ is equal to the square on Α∠.
κοινὸν προσκείσθω τὸ ἀπὸ ΒΓ· ὅλον ἄρα τὸ ἀπὸ ΑΒ τετράγωνον ἴσον ἐστὶ τοῖς ἀπὸ τῶν Α∠, ΓΒ τετραγώνοις.
Let the common square on ΒΓ be added; therefore the whole square on ΑΒ is equal to the squares on Α∠, ΓΒ.
§6.25κε΄.
25.
Ἔστω τὸ ὑπὸ τῶν ΑΒΓ ἴσον τῷ ἀπὸ τῆς Β∠· ὅτι γίνεται γ, τὸ μὲν ὑπὸ τῆς τῶν Α∠, ∠Γ ὑπεροχῆς καὶ τῆς Β∠ ἴσον τῷ ὑπὸ Α∠Γ τὸ δὲ ὑπὸ τῆς τῶν Α∠Γ ὑπεροχῆς καὶ τῆς ΒΓ ἴσον τῷ ἀπὸ τῆς ∠Γ τετραγώνῳ, τὸ δὲ ὑπὸ τῆς τῶν Α∠, ∠Γ ὑπεροχῆς καὶ τῆς ΒΑ ἴσον τῷ ἀπὸ τῆς Α∠ τετραγώνῳ.
Let the rectangle contained by ΑΒ, ΒΓ be equal to the square on Β∠; (to prove) that three things follow: first, the rectangle contained by the difference between Α∠, ∠Γ and Β∠ is equal to the rectangle contained by Α∠, ∠Γ; second, the rectangle contained by the difference between Α∠, ∠Γ and ΒΓ is equal to the square on ∠Γ; third, the rectangle contained by the difference between Α∠, ∠Γ and ΒΑ is equal to the square on Α∠.
ἐπεὶ γάρ ἐστιν, ὡς ἡ ΑΒ πρὸς τὴν Β∠, οὕτως ἡ Β∠ πρὸς τὴν ΒΓ, λοιπὴ πρὸς λοιπὴν καὶ διελόντι· ἔστιν οὖν, ὡς ἡ τῶν Α∠, ∠Γ ὑπεροχὴ πρὸς τὴν ∠Γ, οὕτως ἡ Α∠ πρὸς τὴν ∠Β τὸ ἄρα ὑπὸ τῆς τῶν Α∠, ∠ ὑπεροχῆς καὶ τῆς ∠Β ἴσον ἐστὶν τῷ ὑπὸ τῶν Α∠, ∠Γ. πάλιν, ἐπεὶ λοιπὴ ἡ Α∠ πρὸς λοιπὴν τὴν ∠Γ ἐστιν, ὡς ἡ ∠Β πρὸς τὴν ΒΓ, διελόντι ἐστίν, ὡς ἡ τῶν Α∠Γ ὑπεροχὴ πρὸς τὴν ∠Γ, οὕτως ἡ ∠Γ πρὸς τὴν ΓΒ·
For since, as ΑΒ is to Β∠, so is Β∠ to ΒΓ, remaining is to remaining, and by subtraction; therefore, as the difference between Α∠, ∠Γ is to ∠Γ, so is Α∠ to ∠Β; therefore, the rectangle contained by the difference between Α∠, ∠ and ∠Β is equal to the rectangle contained by Α∠, ∠Γ.
τὸ ἄρα ὑπὸ τῆς τῶν Α∠, ∠Γ ὑπεροχῆς καὶ τῆς ΒΓ ἴσον ἐστὶν τῷ ἀπὸ τῆς ΔΓ τετραγώνῳ.
Again, since the remaining Α∠ is to the remaining ∠Γ, as ∠Β is to ΒΓ, by subtraction, as the difference between Α∠, ∠Γ is to ∠Γ, so is ∠Γ to ΓΒ; therefore, the rectangle contained by the difference between Α∠, ∠Γ and ΒΓ is equal to the square on ΔΓ.
πάλιν, ἐπεί ἐστιν, ὡς ἡ Α∠ πρὸς τὴν ∠Γ οὕτως ἡ ΑΒ πρὸς τὴν Β∠, ἀνάπαλιν καὶ διελόντι ἐστίν, ὡς ἡ τῶν Απαλιν, ∠Γ ὑπεροχὴ πρὸς τὴν ∠Α, οὕτως ἡ ∠Α πρὸς τὴν ΑΒ τὸ ἄρα ὑπὸ τῆς τῶν Α∠, ∠Γ ὑπεροχῆς καὶ τῆς ΑΒ ἴσον ἐστὶν τῷ ἀπὸ τῆς Α∠ τετραγώνῳ.
Again, since, as Α∠ is to ∠Γ, so is ΑΒ to Β∠, by inversion and subtraction, as the difference between Απαλιν, ∠Γ is to ∠Α, so is ∠Α to ΑΒ; therefore, the rectangle contained by the difference between Α∠, ∠Γ and ΑΒ is equal to the square on Α∠.