§10.prop2.63τὸ ἀπὸ τῆς μείζονος παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων τετάρτην.
The square on a major straight line applied to a rational straight line produces as breadth a fourth binomial straight line.
ἔστω μείζων ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, ὥστε μείζονα εἶναι τὴν ΑΓ τῆς ΓΒ, ῥητὴ δὲ ἡ ΔΕ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΔΕ παραβεβλήσθω τὸ ΔΖ παραλληλόγραμμον πλάτος ποιοῦν τὴν ΔΗ· λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ τετάρτη.
Let AB be a major straight line divided at C, so that AC is greater than CB; and let DE be a rational straight line, and let there be applied to DE the parallelogram DZ equal to the square on AB, producing DH as breadth; I say that DH is a fourth binomial straight line.
Κατεσκευάσθω τὰ αὐτὰ τοῖς προδεδειγμένοις.
For let the same construction be made as in what was proved before.
καὶ ἐπεὶ μείζων ἐστὶν ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων ῥητόν, τὸ δὲ ὑπʼ αὐτῶν μέσον.
And since AB is a major straight line divided at C, therefore AC, CB are incommensurable in square, making the sum of the squares on them rational, but the rectangle contained by them medial.
ἐπεὶ οὖν ῥητόν ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, ῥητὸν ἄρα ἐστὶ τὸ ΔΛ· ῥητὴ ἄρα καὶ ἡ ΔΜ καὶ σύμμετρος τῇ ΔΕ μήκει.
Since then the sum of the squares on AC, CB is rational, therefore DL is also rational; therefore DM is also rational and commensurable in length with DE.
πάλιν, ἐπεὶ μέσον ἐστὶ τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, τουτέστι τὸ ΜΖ, καὶ παρὰ ῥητήν ἐστι τὴν ΜΛ, ῥητὴ ἄρα ἐστὶ καὶ ἡ ΜΗ καὶ ἀσύμμετρος τῇ ΔΕ μήκει· ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΔΜ τῇ ΜΗ μήκει.
Again, since twice the rectangle contained by AC, CB, that is, MZ, is medial, and it is applied to the rational straight line ML, therefore MH is also rational and incommensurable in length with DE; therefore DM is also incommensurable in length with MH.
αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ.
δεικτέον, ὅτι καὶ τετάρτη.
Therefore DM, MH are rational straight lines commensurable in square only; therefore DH is a binomial straight line. We must then prove that it is also a fourth binomial straight line.
ὁμοίως δὴ δείξομεν τοῖς πρότερον, ὅτι μείζων ἐστὶν ἡ ΔΜ τῆς ΜΗ, καὶ ὅτι τὸ ὑπὸ ΔΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΜΝ. ἐπεὶ οὖν ἀσύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΓ τῷ ἀπὸ τῆς ΓΒ, ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ΔΘ τῷ ΚΛ· ὥστε ἀσύμμετρος καὶ ἡ ΔΚ τῇ ΚΜ ἐστιν.
Similarly indeed to the prior cases we will prove that DM is greater than MH, and that the rectangle contained by DKM is equal to the square on MN. Since then the square on AC is incommensurable with the square on CB, therefore DΘ is also incommensurable with KL; so that DK is also incommensurable with KM.
ἐὰν δὲ ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παραλληλόγραμμον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ καὶ εἰς ἀσύμμετρα αὐτὴν διαιρῇ, ἡ μείζων τῆς ἐλάσσονος μεῖζον δυνήσεται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει· ἡ ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ.
And 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, deficient by a square figure, and dividing it into incommensurable segments, the square on the greater is greater than the square on the less by the square on a straight line incommensurable in length with the greater; therefore the square on DM is greater than the square on MH by the square on a straight line incommensurable with DM.
καί εἰσιν αἱ ΔΜ, ΜΗ ῥηταὶ δυνάμει μόνον σύμμετροι, καὶ ἡ ΔΜ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΔΕ.
ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ τετάρτη· ὅπερ ἔδει δεῖξαι.
And DM, MH are rational straight lines commensurable in square only, and DM is commensurable with the set out rational straight line DE. Therefore DH is a fourth binomial straight line; which was to be proved.
§10.prop2.64τὸ ἀπὸ τῆς ῥητὸν καὶ μέσον δυναμένης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων πέμπτην.
The square on a straight line which produces with a rational area a medial area applied to a rational straight line produces as breadth a fifth binomial straight line.
ἔστω ῥητὸν καὶ μέσον δυναμένη ἡ ΑΒ διῃρημένη εἰς τὰς εὐθείας κατὰ τὸ Γ, ὥστε μείζονα εἶναι τὴν ΑΓ, καὶ ἐκκείσθω ῥητὴ ἡ ΔΕ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΔΕ παραβεβλήσθω τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ· λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ πέμπτη.
Let AB be a straight line which produces with a rational area a medial area divided into straight lines at C, so that the greater is AC; and let the rational straight line DE be set out, and let there be applied to DE the parallelogram DZ equal to the square on AB, producing DH as breadth; I say that DH is a fifth binomial straight line.
Κατεσκευάσθω τὰ αὐτὰ τοῖς πρὸ τούτου.
For let the same construction be made as in what was proved before.
ἐπεὶ οὖν ῥητὸν καὶ μέσον δυναμένη ἐστὶν ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων μέσον, τὸ δʼ ὑπʼ αὐτῶν ῥητόν.
Since then AB is a straight line which produces with a rational area a medial area divided at C, therefore AC, CB are incommensurable in square, making the sum of the squares on them medial, but the rectangle contained by them rational.
ἐπεὶ οὖν μέσον ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, μέσον ἄρα ἐστὶ τὸ ΔΛ· ὥστε ῥητή ἐστιν ἡ ΔΜ καὶ μήκει ἀσύμμετρος τῇ ΔΕ. πάλιν, ἐπεὶ ῥητόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΓΒ, τουτέστι τὸ ΜΖ, ῥητὴ ἄρα ἡ ΜΗ καὶ σύμμετρος τῇ ΔΕ. ἀσύμμετρος ἄρα ἡ ΔΜ τῇ ΜΗ·
Since then the sum of the squares on AC, CB is medial, therefore DL is also medial; so that DM is rational and incommensurable in length with DE. Again, since twice the rectangle contained by ACB, that is, MZ, is rational, therefore MH is also rational and commensurable with DE.
αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ.
λέγω δή, ὅτι καὶ πέμπτη.
Therefore DM is incommensurable with MH; therefore DM, MH are rational straight lines commensurable in square only; therefore DH is a binomial straight line. We must then prove that it is also a fifth binomial straight line.
ὁμοίως γὰρ δειχθήσεται, ὅτι τὸ ὑπὸ τῶν ΔΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΜΝ, καὶ ἀσύμμετρος ἡ ΔΚ τῇ ΚΜ μήκει· ἡ ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ.
For it will similarly be proved that the rectangle contained by DKM is equal to the square on MN, and that DK is incommensurable in length with KM; therefore the square on DM is greater than the square on MH by the square on a straight line incommensurable with DM.
καί εἰσιν αἱ ΔΜ, ΜΗ δυνάμει μόνον σύμμετροι, καὶ ἡ ἐλάσσων ἡ ΜΗ σύμμετρος τῇ ΔΕ μήκει.
And DM, MH are commensurable in square only, and the less segment MH is commensurable in length with DE.
ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ πέμπτη· ὅπερ ἔδει δεῖξαι.
Therefore DH is a fifth binomial straight line; which was to be proved.