§10.prop1.17ἐὰν ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ καὶ εἰς σύμμετρα αὐτὴν διαιρῇ μήκει, ἡ μείζων τῆς ἐλάσσονος μεῖζον δυνήσεται τῷ ἀπὸ συμμέτρου ἑαυτῇ.
If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and falling short by a square figure, and if it divide it into parts which are commensurable in length, the greater will be greater in square than the less by the square on a straight line commensurable with itself.
καὶ ἐὰν ἡ μείζων τῆς ἐλάσσονος μεῖζον δύνηται τῷ ἀπὸ συμμέτρου ἑαυτῇ, τῷ δὲ τετάρτῳ τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς σύμμετρα αὐτὴν διαιρεῖ μήκει.
And if the greater be greater in square than the less by the square on a straight line commensurable with itself, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and falling short by a square figure, it divides it into parts which are commensurable in length.
ἔστωσαν δύο εὐθεῖαι ἄνισοι αἱ Α, ΒΓ, ὧν μείζων ἡ ΒΓ, τῷ δὲ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ἐλάσσονος τῆς Α, τουτέστι τῷ ἀπὸ τῆς ἡμισείας τῆς Α, ἴσον παρὰ τὴν ΒΓ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΒΔ, ΔΓ, σύμμετρος δὲ ἔστω ἡ ΒΔ τῇ ΔΓ μήκει·
Let there be two unequal straight lines A, BΓ, of which BΓ is the greater, and to BΓ let there be applied a parallelogram equal to the fourth part of the square on the less A, that is, to the square on the half of A, falling short by a square figure, and let it be the rectangle contained by BΔ, ΔΓ; and let BΔ be commensurable in length with ΔΓ.
λέγω, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ.
I say that the square on BΓ is greater than the square on A by the square on a straight line commensurable in length with itself.
τετμήσθω γὰρ ἡ ΒΓ δίχα κατὰ τὸ Ε σημεῖον, καὶ κείσθω τῇ ΔΕ ἴση ἡ ΕΖ. λοιπὴ ἄρα ἡ ΔΓ ἴση ἐστὶ τῇ ΒΖ. καὶ ἐπεὶ εὐθεῖα ἡ ΒΓ τέτμηται εἰς μὲν ἴσα κατὰ τὸ Ε, εἰς δὲ ἄνισα κατὰ τὸ Δ, τὸ ἄρα ὑπὸ ΒΔ, ΔΓ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΕΔ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΓ τετραγώνῳ·
For let BΓ be bisected at the point E, and let EZ be made equal to ΔE; therefore the remainder ΔΓ is equal to BZ. And since the straight line BΓ has been cut into equal parts at E, and into unequal parts at Δ, therefore the rectangle contained by BΔ, ΔΓ together with the square on EΔ is equal to the square on EΓ.
καὶ τὰ τετραπλάσια· τὸ ἄρα τετράκις ὑπὸ τῶν ΒΔ, ΔΓ μετὰ τοῦ τετραπλασίου τοῦ ἀπὸ τῆς ΔΕ ἴσον ἐστὶ τῷ τετράκις ἀπὸ τῆς ΕΓ τετραγώνῳ.
And the quadruples of them are also equal; therefore four times the rectangle contained by BΔ, ΔΓ together with four times the square on ΔE is equal to four times the square on EΓ.
ἀλλὰ τῷ μέν τετραπλασίῳ τοῦ ὑπὸ τῶν ΒΔ, ΔΓ ἴσον ἐστὶ τὸ ἀπὸ τῆς Α τετράγωνον, τῷ δὲ τετραπλασίῳ τοῦ ἀπὸ τῆς ΔΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΔΖ τετράγωνον· διπλασίων γάρ ἐστιν ἡ ΔΖ τῆς ΔΕ. τῷ δὲ τετραπλασίῳ τοῦ ἀπὸ τῆς ΕΓ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΒΓ τετράγωνον·
But the square on A is equal to four times the rectangle contained by BΔ, ΔΓ; and the square on ΔZ is equal to four times the square on ΔE, for ΔZ is double of ΔE; and the square on BΓ is equal to four times the square on EΓ, for BΓ is again double of ΓE.
διπλασίων γάρ ἐστι πάλιν ἡ ΒΓ τῆς ΓΕ. τὰ ἄρα ἀπὸ τῶν Α, ΔΖ τετράγωνα ἴσα ἐστὶ τῷ ἀπὸ τῆς ΒΓ τετραγώνῳ· ὥστε τὸ ἀπὸ τῆς ΒΓ τοῦ ἀπὸ τῆς Α μεῖζόν ἐστι τῷ ἀπὸ τῆς ΔΖ·
Therefore the squares on A, ΔZ are equal to the square on BΓ; so that the square on BΓ is greater than the square on A by the square on ΔZ; therefore the square on BΓ is greater than the square on A by ΔZ.
ἡ ΒΓ ἄρα τῆς Α μεῖζον δύναται τῇ ΔΖ. δεικτέον, ὅτι καὶ σύμμετρός ἐστιν ἡ ΒΓ τῇ ΔΖ. ἐπεὶ γὰρ σύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ μήκει, σύμμετρος ἄρα ἐστὶ καὶ ἡ ΒΓ τῇ ΓΔ μήκει.
It must be shown that BΓ is also commensurable in length with ΔZ. For since BΔ is commensurable in length with ΔΓ, therefore BΓ is also commensurable in length with ΓΔ.
ἀλλὰ ἡ ΓΔ ταῖς ΓΔ, ΒΖ ἐστι σύμμετρος μήκει·
But ΓΔ is commensurable in length with the sum of ΓΔ, BZ; for ΓΔ is equal to BZ.
ἴση γάρ ἐστιν ἡ ΓΔ τῇ ΒΖ. καὶ ἡ ΒΓ ἄρα σύμμετρός ἐστι ταῖς ΒΖ, ΓΔ μήκει· ὥστε καὶ λοιπῇ τῇ ΖΔ σύμμετρός ἐστιν ἡ ΒΓ μήκει·
Therefore BΓ is also commensurable in length with the sum of BZ, ΓΔ; so that BΓ is also commensurable in length with the remainder ZΔ.
ἡ ΒΓ ἄρα τῆς Α μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ.
Therefore the square on BΓ is greater than the square on A by the square on a straight line commensurable in length with itself.
ἀλλὰ δὴ ἡ ΒΓ τῆς Α μεῖζον δυνάσθω τῷ ἀπὸ συμμέτρου ἑαυτῇ, τῷ δὲ τετάρτῳ τοῦ ἀπὸ τῆς Α ἴσον παρὰ τὴν ΒΓ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΒΔ, ΔΓ. δεικτέον, ὅτι σύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ μήκει.
Next, let the square on BΓ be greater than the square on A by the square on a straight line commensurable in length with itself, and to BΓ let there be applied a parallelogram equal to the fourth part of the square on A and falling short by a square figure, and let it be the rectangle contained by BΔ, ΔΓ. It must be shown that BΔ is commensurable in length with ΔΓ.
τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ. δύναται δὲ ἡ ΒΓ τῆς Α μεῖζον τῷ ἀπὸ συμμέτρου ἑαυτῇ. σύμμετρος ἄρα ἐστὶν ἡ ΒΓ τῇ ΖΔ μήκει·
For, with the same construction, we shall similarly show that the square on BΓ is greater than the square on A by the square on ZΔ. But the square on BΓ is greater than the square on A by the square on a straight line commensurable with itself; therefore BΓ is commensurable in length with ZΔ.
ὥστε καὶ λοιπῇ συναμφοτέρῳ τῇ ΒΖ, ΔΓ σύμμετρός ἐστιν ἡ ΒΓ μήκει.
So that BΓ is also commensurable in length with the remainder, the sum of BZ, ΔΓ.
ἀλλὰ συναμφότερος ἡ ΒΖ, ΔΓ σύμμετρός ἐστι τῇ ΔΓ. ὥστε καὶ ἡ ΒΓ τῇ ΓΔ σύμμετρός ἐστι μήκει·
But the sum of BZ, ΔΓ is commensurable with ΔΓ; so that BΓ is also commensurable in length with ΓΔ.
καὶ διελόντι ἄρα ἡ ΒΔ τῇ ΔΓ ἐστι σύμμετρος μήκει.
And therefore, by separation, BΔ is commensurable in length with ΔΓ.
ἐὰν ἄρα ὦσι δύο εὐθεῖαι ἄνισοι, καὶ τὰ ἑξῆς.
Therefore, if there be two unequal straight lines, and the rest.