§10.prop1.12τὰ τῷ αὐτῷ μεγέθει σύμμετρα καὶ ἀλλήλοις ἐστὶ σύμμετρα.
Magnitudes commensurable with the same magnitude are also commensurable with one another.
ἑκάτερον γὰρ τῶν Α, Β τῷ Γ ἔστω σύμμετρον. λέγω, ὅτι καὶ τὸ Α τῷ Β ἐστι σύμμετρον.
For let each of the magnitudes A, B be commensurable with Γ; I say that A is also commensurable with B.
ἐπεὶ γὰρ σύμμετρόν ἐστι τὸ Α τῷ Γ, τὸ Α ἄρα πρὸς τὸ Γ λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν.
For since A is commensurable with Γ, therefore A has to Γ the ratio which a number has to a number.
ἐχέτω, ὃν ὁ Δ πρὸς τὸν Ε. πάλιν, ἐπεὶ σύμμετρόν ἐστι τὸ Γ τῷ Β, τὸ Γ ἄρα πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν.
Let it have that which Δ has to E. Again, since Γ is commensurable with B, therefore Γ has to B the ratio which a number has to a number.
ἐχέτω, ὃν ὁ Ζ πρὸς τὸν Η. καὶ λόγων δοθέντων ὁποσωνοῦν τοῦ τε, ὃν ἔχει ὁ Δ πρὸς τὸν Ε, καὶ ὁ Ζ πρὸς τὸν Η εἰλήφθωσαν ἀριθμοὶ ἑξῆς ἐν τοῖς δοθεῖσι λόγοις οἱ Θ, Κ, Λ·
Let it have that which Z has to H. And, any number of ratios being given, namely that which Δ has to E, and that which Z has to H, let numbers Θ, K, Λ be taken continuously in the given ratios, so that, as Δ is to E, so is Θ to K, and as Z is to H, so is K to Λ.
ὥστε εἶναι ὡς μὲν τὸν Δ πρὸς τὸν Ε, οὕτως τὸν Θ πρὸς τὸν Κ, ὡς δὲ τὸν Ζ πρὸς τὸν Η, οὕτως τὸν Κ πρὸς τὸν Λ.
ἐπεὶ οὖν ἐστιν ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ πρὸς τὸν Ε, ἀλλʼ ὡς ὁ Δ πρὸς τὸν Ε, οὕτως ὁ Θ πρὸς τὸν Κ, ἔστιν ἄρα καὶ ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Θ πρὸς τὸν Κ. πάλιν, ἐπεί ἐστιν ὡς τὸ Γ πρὸς τὸ Β, οὕτως ὁ Ζ πρὸς τὸν Η, ἀλλʼ ὡς ὁ Ζ πρὸς τὸν Η, ὁ Κ πρὸς τὸν Λ, καὶ ὡς ἄρα τὸ Γ πρὸς τὸ Β, οὕτως ὁ Κ πρὸς τὸν Λ. ἔστι δὲ καὶ ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Θ πρὸς τὸν Κ· διʼ ἴσου ἄρα ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως ὁ Θ πρὸς τὸν Λ. τὸ Α ἄρα πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς ὁ Θ πρὸς ἀριθμὸν τὸν Λ· σύμμετρον ἄρα ἐστὶ τὸ Α τῷ Β.
τὰ ἄρα τῷ αὐτῷ μεγέθει σύμμετρα καὶ ἀλλήλοις ἐστὶ σύμμετρα· ὅπερ ἔδει δεῖξαι.
Since, then, as A is to Γ, so is Δ to E, but as Δ is to E, so is Θ to K, therefore also, as A is to Γ, so is Θ to K. Again, since as Γ is to B, so is Z to H, but as Z is to H, so is K to Λ, therefore also, as Γ is to B, so is K to Λ. And it is also, as A is to Γ, so is Θ to K; therefore, ex aequali, as A is to B, so is Θ to Λ. Therefore A has to B the ratio which the number Θ has to the number Λ; therefore A is commensurable with B. Therefore magnitudes commensurable with the same magnitude are also commensurable with one another; which was to be proved.
§10.prop1.13ἐὰν ᾖ δύο μεγέθη σύμμετρα, τὸ δὲ ἕτερον αὐτῶν μεγέθει τινὶ ἀσύμμετρον ᾖ, καὶ τὸ λοιπὸν τῷ αὐτῷ ἀσύμμετρον ἔσται.
If two magnitudes be commensurable, and one of them be incommensurable with any magnitude, the remaining one will also be incommensurable with the same.
ἔστω δύο μεγέθη σύμμετρα τὰ Α, Β, τὸ δὲ ἕτερον αὐτῶν τὸ Α ἄλλῳ τινὶ τῷ Γ ἀσύμμετρον ἔστω· λέγω, ὅτι καὶ τὸ λοιπὸν τὸ Β τῷ Γ ἀσύμμετρόν ἐστιν.
Let two commensurable magnitudes be A, B, and let one of them, A, be incommensurable with any other magnitude Γ; I say that the remaining one, B, is also incommensurable with Γ.
εἰ γάρ ἐστι σύμμετρον τὸ Β τῷ Γ, ἀλλὰ καὶ τὸ Α τῷ Β σύμμετρόν ἐστιν, καὶ τὸ Α ἄρα τῷ Γ σύμμετρόν ἐστιν.
For if B is commensurable with Γ, since A is also commensurable with B, A is therefore also commensurable with Γ.
ἀλλὰ καὶ ἀσύμμετρον· ὅπερ ἀδύνατον.
But it is also incommensurable; which is impossible.
οὐκ ἄρα σύμμετρόν ἐστι τὸ Β τῷ Γ· ἀσύμμετρον ἄρα.
Therefore B is not commensurable with Γ; therefore it is incommensurable.
ἐὰν ἄρα ᾖ δύο μεγέθη σύμμετρα, καὶ τὰ ἑξῆς.
Therefore, if two magnitudes be commensurable, and the rest.
λῆμμα
δύο δοθεισῶν εὐθειῶν ἀνίσων εὑρεῖν, τίνι μεῖζον δύναται ἡ μείζων τῆς ἐλάσσονος.
Lemma To find, when two unequal straight lines are given, by how much the square on the greater is greater than that on the less.
ἔστωσαν αἱ δοθεῖσαι δύο ἄνισοι εὐθεῖαι αἱ ΑΒ, Γ, ὧν μείζων ἔστω ἡ ΑΒ· δεῖ δὴ εὑρεῖν, τίνι μεῖζον δύναται ἡ ΑΒ τῆς Γ.
γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ εἰς αὐτὸ ἐνηρμόσθω τῇ Γ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΔΒ. φανερὸν δή, ὅτι ὀρθή ἐστιν ἡ ὑπὸ ΑΔΒ γωνία, καὶ ὅτι ἡ ΑΒ τῆς ΑΔ, τουτέστι τῆς Γ, μεῖζον δύναται τῇ ΔΒ.
ὁμοίως δὲ καὶ δύο δοθεισῶν εὐθειῶν ἡ δυναμένη αὐτὰς εὑρίσκεται οὕτως. ἔστωσαν αἱ δοθεῖσαι δύο εὐθεῖαι αἱ ΑΔ, ΔΒ, καὶ δέον ἔστω εὑρεῖν τὴν δυναμένην αὐτάς. κείσθωσαν γάρ, ὥστε ὀρθὴν γωνίαν περιέχειν τὴν ὑπὸ ΑΔ, ΔΒ, καὶ ἐπεζεύχθω ἡ ΑΒ· φανερὸν πάλιν, ὅτι ἡ τὰς ΑΔ, ΔΒ δυναμένη ἐστὶν ἡ ΑΒ·
Let the given two unequal straight lines be AB, Γ, of which let AB be the greater; it is indeed required to find by how much the square on AB is greater than that on Γ. Let the semicircle ADB be described on AB, and let AD equal to Γ be fitted into it, and let ΔB be joined. It is then manifest that the angle ADB is a right angle, and that the square on AB is greater than that on AD, that is, Γ, by the square on ΔB. And in like manner, when two straight lines are given, the straight line equal in square to them is also found thus. Let the given two straight lines be AD, ΔB, and let it be required to find the straight line equal in square to them. For let them be placed so as to contain a right angle contained by AD, ΔB, and let AB be joined.
ὅπερ ἔδει δεῖξαι.
It is then again manifest that the straight line equal in square to AD, ΔB is AB; which was to be proved.