§10.prop1.19τὸ ὑπὸ ῥητῶν μήκει συμμέτρων κατά τινα τῶν προειρημένων τρόπων εὐθειῶν περιεχόμενον ὀρθογώνιον ῥητόν ἐστιν.
The rectangle contained by rational straight lines commensurable in length in any of the aforesaid ways is rational.
ὑπὸ γὰρ ῥητῶν μήκει συμμέτρων εὐθειῶν τῶν ΑΒ, ΒΓ ὀρθογώνιον περιεχέσθω τὸ ΑΓ· λέγω, ὅτι ῥητόν ἐστι τὸ ΑΓ.
Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔ·
For let the rectangle AΓ be contained by rational straight lines AB, BΓ commensurable in length; I say that AΓ is rational.
ῥητὸν ἄρα ἐστὶ τὸ ΑΔ. καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει, ἴση δέ ἐστιν ἡ ΑΒ τῇ ΒΔ, σύμμετρος ἄρα ἐστὶν ἡ ΒΔ τῇ ΒΓ μήκει.
For let there be described on AB the square AΔ; therefore AΔ is rational. And since AB is commensurable with BΓ in length, and AB is equal to BΔ, therefore BΔ is also commensurable with BΓ in length.
καί ἐστιν ὡς ἡ ΒΔ πρὸς τὴν ΒΓ, οὕτως τὸ ΔΑ πρὸς τὸ ΑΓ. σύμμετρον ἄρα ἐστὶ τὸ ΔΑ τῷ ΑΓ. ῥητὸν δὲ τὸ ΔΑ· ῥητὸν ἄρα ἐστὶ καὶ τὸ ΑΓ.
τὸ ἄρα ὑπὸ ῥητῶν μήκει συμμέτρων, καὶ τὰ ἑξῆς.
And as BΔ is to BΓ, so is ΔA to AΓ. Therefore ΔA is commensurable with AΓ. But ΔA is rational; therefore AΓ is also rational. Therefore the rectangle contained by rational straight lines commensurable in length, and the rest.
§10.prop1.20ἐὰν ῥητὸν παρὰ ῥητὴν παραβληθῇ, πλάτος ποιεῖ ῥητὴν καὶ σύμμετρον τῇ, παρʼ ἣν παράκειται, μήκει.
If a rational area be applied to a rational straight line, it produces as breadth a straight line which is rational and commensurable in length with the straight line to which it is applied.
ῥητὸν γὰρ τὸ ΑΓ παρὰ ῥητὴν κατά τινα πάλιν τῶν προειρημένων τρόπων τὴν ΑΒ παραβεβλήσθω πλάτος ποιοῦν τὴν ΒΓ· λέγω, ὅτι ῥητή ἐστιν ἡ ΒΓ καὶ σύμμετρος τῇ ΒΑ μήκει.
For let the rational area AΓ be applied to the straight line AB, which is rational in any of the aforesaid ways, producing as breadth BΓ; I say that BΓ is rational and commensurable in length with BA.
Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔ· ῥητὸν ἄρα ἐστὶ τὸ ΑΔ. ῥητὸν δὲ καὶ τὸ ΑΓ· σύμμετρον ἄρα ἐστὶ τὸ ΔΑ τῷ ΑΓ. καί ἐστιν ὡς τὸ ΔΑ πρὸς τὸ ΑΓ, οὕτως ἡ ΔΒ πρὸς τὴν ΒΓ. σύμμετρος ἄρα ἐστὶ καὶ ἡ ΔΒ τῇ ΒΓ·
For let there be described on AB the square AΔ; therefore AΔ is rational. But AΓ is also rational; therefore ΔA is commensurable with AΓ. And as ΔA is to AΓ, so is ΔB to BΓ. Therefore ΔB is also commensurable with BΓ.
ἴση δὲ ἡ ΔΒ τῇ ΒΑ· σύμμετρος ἄρα καὶ ἡ ΑΒ τῇ ΒΓ. ῥητὴ δέ ἐστιν ἡ ΑΒ·
But ΔB is equal to BA; therefore AB is also commensurable with BΓ.
ῥητὴ ἄρα ἐστὶ καὶ ἡ ΒΓ καὶ σύμμετρος τῇ ΑΒ μήκει.
And AB is rational; therefore BΓ is also rational and commensurable in length with AB.
ἐὰν ἄρα ῥητὸν παρὰ ῥητὴν παραβληθῇ, καὶ τὰ ἑξῆς.
Therefore, if a rational area be applied to a rational straight line, and the rest.
§10.prop1.21τὸ ὑπὸ ῥητῶν δυνάμει μόνον συμμέτρων εὐθειῶν περιεχόμενον ὀρθογώνιον ἄλογόν ἐστιν, καὶ ἡ δυναμένη αὐτὸ ἄλογός ἐστιν, καλείσθω δὲ μέση.
The rectangle contained by rational straight lines commensurable in square only is irrational, and the side of the square equal to it is irrational, and let it be called medial.
ὑπὸ γὰρ ῥητῶν δυνάμει μόνον συμμέτρων εὐθειῶν τῶν ΑΒ, ΒΓ ὀρθογώνιον περιεχέσθω τὸ ΑΓ· λέγω, ὅτι ἄλογόν ἐστι τὸ ΑΓ, καὶ ἡ δυναμένη αὐτὸ ἄλογός ἐστιν, καλείσθω δὲ μέση.
For let the rectangle AΓ be contained by rational straight lines AB, BΓ commensurable in square only; I say that AΓ is irrational, and the side of the square equal to it is irrational, and let it be called medial.
Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔ·
For let there be described on AB the square AΔ; therefore AΔ is rational.
ῥητὸν ἄρα ἐστὶ τὸ ΑΔ. καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει· δυνάμει γὰρ μόνον ὑπόκεινται σύμμετροι· ἴση δὲ ἡ ΑΒ τῇ ΒΔ, ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΔΒ τῇ ΒΓ μήκει.
And since AB is incommensurable in length with BΓ (for they are assumed to be commensurable in square only), and AB is equal to BΔ, therefore ΔB is also incommensurable in length with BΓ.
καί ἐστιν ὡς ἡ ΔΒ πρὸς τὴν ΒΓ, οὕτως τὸ ΑΔ πρὸς τὸ ΑΓ· ἀσύμμετρον ἄρα τὸ ΔΑ τῷ ΑΓ. ῥητὸν δὲ τὸ ΔΑ· ἄλογον ἄρα ἐστὶ τὸ ΑΓ·
And as ΔB is to BΓ, so is AΔ to AΓ; therefore ΔA is incommensurable with AΓ. But ΔA is rational; therefore AΓ is irrational.
ὥστε καὶ ἡ δυναμένη τὸ ΑΓ ἄλογός ἐστιν, καλείσθω δὲ μέση· ὅπερ ἔδει δεῖξαι.
So that the side of the square equal to AΓ is also irrational, and let it be called medial; which was to be proved.
λῆμμα
ἐὰν ὦσι δύο εὐθεῖαι, ἔστιν ὡς ἡ πρώτη πρὸς τὴν δευτέραν, οὕτως τὸ ἀπὸ τῆς πρώτης πρὸς τὸ ὑπὸ τῶν δύο εὐθειῶν.
Lemma If there be two straight lines, as the first is to the second, so is the square on the first to the rectangle contained by the two straight lines.
ἔστωσαν δύο εὐθεῖαι αἱ ΖΕ, ΕΗ. λέγω, ὅτι ἐστὶν ὡς ἡ ΖΕ πρὸς τὴν ΕΗ, οὕτως τὸ ἀπὸ τῆς ΖΕ πρὸς τὸ ὑπὸ τῶν ΖΕ, ΕΗ.
Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΖΕ τετράγωνον τὸ ΔΖ, καὶ συμπεπληρώσθω τὸ ΗΔ. ἐπεὶ οὖν ἐστιν ὡς ἡ ΖΕ πρὸς τὴν ΕΗ, οὕτως τὸ ΖΔ πρὸς τὸ ΔΗ, καί ἐστι τὸ μὲν ΖΔ τὸ ἀπὸ τῆς ΖΕ, τὸ δὲ ΔΗ τὸ ὑπὸ τῶν ΔΕ, ΕΗ, τουτέστι τὸ ὑπὸ τῶν ΖΕ, ΕΗ, ἔστιν ἄρα ὡς ἡ ΖΕ τὴν ΕΗ, οὕτως τὸ ἀπὸ τῆς ΖΕ πρὸς τὸ ὑπὸ τῶν ΖΕ, ΕΗ. ὁμοίως δὲ καὶ ὡς τὸ ὑπὸ τῶν ΗΕ, ΕΖ πρὸς τὸ ἀπὸ τῆς ΕΖ, τουτέστιν ὡς τὸ ΗΔ πρὸς τὸ ΖΔ, οὕτως ἡ ΗΕ πρὸς τὴν ΕΖ· ὅπερ ἔδει δεῖξαι.
Let there be two straight lines ZE, EH. I say that, as ZE is to EH, so is the square on ZE to the rectangle contained by ZE, EH. For let there be described on ZE the square ΔZ, and let the rectangle HΔ be completed. Since then as ZE is to EH, so is ZΔ to ΔH, and ZΔ is the square on ZE, while ΔH is the rectangle contained by ΔE, EH, that is, the rectangle contained by ZE, EH, therefore, as ZE is to EH, so is the square on ZE to the rectangle contained by ZE, EH. Similarly also, as the rectangle contained by HE, EZ is to the square on EZ, that is, as HΔ is to ZΔ, so is HE to EZ; which was to be proved.