§10.prop1.15ἐὰν δύο μεγέθη σύμμετρα συντεθῇ, καὶ τὸ ὅλον ἑκατέρῳ αὐτῶν σύμμετρον ἔσται· κἂν τὸ ὅλον ἑνὶ αὐτῶν σύμμετρον ᾖ, καὶ τὰ ἐξ ἀρχῆς μεγέθη σύμμετρα ἔσται.
If two commensurable magnitudes be added together, the whole will also be commensurable with each of them; and if the whole be commensurable with one of them, the original magnitudes will also be commensurable.
Συγκείσθω γὰρ δύο μεγέθη σύμμετρα τὰ ΑΒ, ΒΓ· λέγω, ὅτι καὶ ὅλον τὸ ΑΓ ἑκατέρῳ τῶν ΑΒ, ΒΓ ἐστι σύμμετρον.
For let two commensurable magnitudes AB, BΓ be added together; I say that the whole AΓ is also commensurable with each of the magnitudes AB, BΓ.
ἐπεὶ γὰρ σύμμετρά ἐστι τὰ ΑΒ, ΒΓ, μετρήσει τι αὐτὰ μέγεθος.
For since AB, BΓ are commensurable, some magnitude will measure them.
μετρείτω, καὶ ἔστω τὸ Δ. ἐπεὶ οὖν τὸ Δ τὰ ΑΒ, ΒΓ μετρεῖ, καὶ ὅλον τὸ ΑΓ μετρήσει.
Let it measure them, and let it be Δ. Since then Δ measures AB, BΓ, it will also measure the whole AΓ.
μετρεῖ δὲ καὶ τὰ ΑΒ, ΒΓ. τὸ Δ ἄρα τὰ ΑΒ, ΒΓ, ΑΓ μετρεῖ·
But it also measures AB, BΓ; therefore Δ measures AB, BΓ, AΓ.
σύμμετρον ἄρα ἐστὶ τὸ ΑΓ ἑκατέρῳ τῶν ΑΒ, ΒΓ.
ἀλλὰ δὴ τὸ ΑΓ ἔστω σύμμετρον τῷ ΑΒ· λέγω δή, ὅτι καὶ τὰ ΑΒ, ΒΓ σύμμετρά ἐστιν.
Therefore AΓ is commensurable with each of the magnitudes AB, BΓ. Next, let AΓ be commensurable with AB; I say indeed that AB, BΓ are also commensurable.
ἐπεὶ γὰρ σύμμετρά ἐστι τὰ ΑΓ, ΑΒ, μετρήσει τι αὐτὰ μέγεθος.
For since AΓ, AB are commensurable, some magnitude will measure them.
μετρείτω, καὶ ἔστω τὸ Δ. ἐπεὶ οὖν τὸ Δ τὰ ΓΑ, ΑΒ μετρεῖ, καὶ λοιπὸν ἄρα τὸ ΒΓ μετρήσει.
Let it measure them, and let it be Δ. Since then Δ measures ΓA, AB, therefore it will also measure the remainder BΓ.
μετρεῖ δὲ καὶ τὸ ΑΒ· τὸ Δ ἄρα τὰ ΑΒ, ΒΓ μετρήσει·
But it also measures AB; therefore Δ will measure AB, BΓ.
σύμμετρα ἄρα ἐστὶ τὰ ΑΒ, ΒΓ.
ἐὰν ἄρα δύο μεγέθη, καὶ τὰ ἑξῆς.
Therefore AB, BΓ are commensurable. Therefore, if two magnitudes, and the rest.
§10.prop1.16ἐὰν δύο μεγέθη ἀσύμμετρα συντεθῇ, καὶ τὸ ὅλον ἑκατέρῳ αὐτῶν ἀσύμμετρον ἔσται· κἂν τὸ ὅλον ἑνὶ αὐτῶν ἀσύμμετρον ᾖ, καὶ τὰ ἐξ ἀρχῆς μεγέθη ἀσύμμετρα ἔσται.
If two incommensurable magnitudes be added together, the whole will also be incommensurable with each of them; and if the whole be incommensurable with one of them, the original magnitudes will also be incommensurable.
Συγκείσθω γὰρ δύο μεγέθη ἀσύμμετρα τὰ ΑΒ, ΒΓ· λέγω, ὅτι καὶ ὅλον τὸ ΑΓ ἑκατέρῳ τῶν ΑΒ, ΒΓ ἀσύμμετρόν ἐστιν.
For let two incommensurable magnitudes AB, BΓ be added together; I say that the whole AΓ is also incommensurable with each of the magnitudes AB, BΓ.
εἰ γὰρ μή ἐστιν ἀσύμμετρα τὰ ΓΑ, ΑΒ, μετρήσει τι μέγεθος.
For if ΓA, AB are not incommensurable, some magnitude will measure them.
μετρείτω, εἰ δυνατόν, καὶ ἔστω τὸ Δ. ἐπεὶ οὖν τὸ Δ τὰ ΓΑ, ΑΒ μετρεῖ, καὶ λοιπὸν ἄρα τὸ ΒΓ μετρήσει.
Let it measure them, if possible, and let it be Δ. Since then Δ measures ΓA, AB, therefore it will also measure the remainder BΓ.
μετρεῖ δὲ καὶ τὸ ΑΒ· τὸ Δ ἄρα τὰ ΑΒ, ΒΓ μετρεῖ.
But it also measures AB; therefore Δ measures AB, BΓ.
σύμμετρα ἄρα ἐστὶ τὰ ΑΒ, ΒΓ·
Therefore AB, BΓ are commensurable.
ὑπέκειντο δὲ καὶ ἀσύμμετρα· ὅπερ ἐστὶν ἀδύνατον.
But they were also assumed to be incommensurable; which is impossible.
οὐκ ἄρα τὰ ΓΑ, ΑΒ μετρήσει τι μέγεθος· ἀσύμμετρα ἄρα ἐστὶ τὰ ΓΑ, ΑΒ. ὁμοίως δὴ δείξομεν, ὅτι καὶ τὰ ΑΓ, ΓΒ ἀσύμμετρά ἐστιν.
Therefore no magnitude will measure ΓA, AB; therefore ΓA, AB are incommensurable. Similarly indeed we shall show that AΓ, ΓB are also incommensurable.
τὸ ΑΓ ἄρα ἑκατέρῳ τῶν ΑΒ, ΒΓ ἀσύμμετρόν ἐστιν.
Therefore AΓ is incommensurable with each of the magnitudes AB, BΓ.
ἀλλὰ δὴ τὸ ΑΓ ἑνὶ τῶν ΑΒ, ΒΓ ἀσύμμετρον ἔστω.
Next, let AΓ be incommensurable with one of the magnitudes AB, BΓ.
ἔστω δὴ πρότερον τῷ ΑΒ· λέγω, ὅτι καὶ τὰ ΑΒ, ΒΓ ἀσύμμετρά ἐστιν.
Let it indeed be first incommensurable with AB; I say that AB, BΓ are also incommensurable.
εἰ γὰρ ἔσται σύμμετρα, μετρήσει τι αὐτὰ μέγεθος.
For if they are commensurable, some magnitude will measure them.
μετρείτω, καὶ ἔστω τὸ Δ. ἐπεὶ οὖν τὸ Δ τὰ ΑΒ, ΒΓ μετρεῖ, καὶ ὅλον ἄρα τὸ ΑΓ μετρήσει.
Let it measure them, and let it be Δ. Since then Δ measures AB, BΓ, therefore it will also measure the whole AΓ.
μετρεῖ δὲ καὶ τὸ ΑΒ· τὸ Δ ἄρα τὰ ΓΑ, ΑΒ μετρεῖ.
But it also measures AB; therefore Δ measures ΓA, AB.
σύμμετρα ἄρα ἐστὶ τὰ ΓΑ, ΑΒ·
Therefore ΓA, AB are commensurable.
ὑπέκειτο δὲ καὶ ἀσύμμετρα· ὅπερ ἐστὶν ἀδύνατον.
But they were also assumed to be incommensurable; which is impossible.
οὐκ ἄρα τὰ ΑΒ, ΒΓ μετρήσει τι μέγεθος· ἀσύμμετρα ἄρα ἐστὶ τὰ ΑΒ, ΒΓ.
ἐὰν ἄρα δύο μεγέθη, καὶ τὰ ἑξῆς. λῆμμα
ἐὰν παρά τινα εὐθεῖαν παραβληθῇ παραλληλόγραμμον ἐλλεῖπον εἴδει τετραγώνῳ, τὸ παραβληθὲν ἴσον ἐστὶ τῷ ὑπὸ τῶν ἐκ τῆς παραβολῆς γενομένων τμημάτων τῆς εὐθείας.
Therefore no magnitude will measure AB, BΓ; therefore AB, BΓ are incommensurable. Therefore, if two magnitudes, and the rest. λῆμμα If a parallelogram falling short by a square figure be applied to a straight line, the applied parallelogram is equal to the rectangle contained by the segments of the straight line which are made by the application.
παρὰ γὰρ εὐθεῖαν τὴν ΑΒ παραβεβλήσθω παραλληλόγραμμον τὸ ΑΔ ἐλλεῖπον εἴδει τετραγώνῳ τῷ ΔΒ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΔ τῷ ὑπὸ τῶν ΑΓ, ΓΒ.
καί ἐστιν αὐτόθεν φανερόν·
For let there be applied to the straight line AB the parallelogram AΔ falling short by a square figure ΔB; I say that AΔ is equal to the rectangle contained by AΓ, ΓB. And this is immediately manifest.
ἐπεὶ γὰρ τετράγωνόν ἐστι τὸ ΔΒ, ἴση ἐστὶν ἡ ΔΓ τῇ ΓΒ, καί ἐστι τὸ ΑΔ τὸ ὑπὸ τῶν ΑΓ, ΓΔ, τουτέστι τὸ ὑπὸ τῶν ΑΓ, ΓΒ.
ἐὰν ἄρα παρά τινα εὐθεῖαν, καὶ τὰ ἑξῆς.
For since ΔB is a square, ΔΓ is equal to ΓB, and AΔ is the rectangle contained by AΓ, ΓΔ, that is, the rectangle contained by AΓ, ΓB. Therefore, if to any straight line, and the rest.