§10.prop2.61τὸ ἀπὸ τῆς ἐκ δύο μέσων πρώτης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων δευτέραν.
The square on a first bimedial straight line applied to a rational straight line produces as breadth a second binomial straight line.
ἔστω ἐκ δύο μέσων πρώτη ἡ ΑΒ διῃρημένη εἰς τὰς μέσας κατὰ τὸ Γ, ὧν μείζων ἡ ΑΓ, καὶ ἐκκείσθω ῥητὴ ἡ ΔΕ, καὶ παραβεβλήσθω παρὰ τὴν ΔΕ τῷ ἀπὸ τῆς ΑΒ ἴσον παραλληλόγραμμον τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ· λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ δευτέρα.
Let AB be a first bimedial straight line divided into its medial terms at C, of which 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 second binomial straight line.
Κατεσκευάσθω γὰρ τὰ αὐτὰ τοῖς πρὸ τούτου.
For let the same construction be made as in the proposition before this.
καὶ ἐπεὶ ἡ ΑΒ ἐκ δύο μέσων ἐστὶ πρώτη διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι ῥητὸν περιέχουσαι· ὥστε καὶ τὰ ἀπὸ τῶν ΑΓ, ΓΒ μέσα ἐστίν.
And since AB is a first bimedial straight line divided at C, therefore AC, CB are medial straight lines commensurable in square only containing a rational rectangle; so that the squares on AC, CB are also medial.
μέσον ἄρα ἐστὶ τὸ ΔΛ. καὶ παρὰ ῥητὴν τὴν ΔΕ παραβέβληται·
Therefore DL is also medial.
ῥητὴ ἄρα ἐστίν ἡ ΜΔ καὶ ἀσύμμετρος τῇ ΔΕ μήκει.
And it has been applied to the rational straight line DE; therefore MD is rational and incommensurable in length with DE.
πάλιν, ἐπεὶ ῥητόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, ῥητόν ἐστι καὶ τὸ ΜΖ. καὶ παρὰ ῥητὴν τὴν ΜΛ παράκειται·
Again, since twice the rectangle contained by AC, CB is rational, MZ is also rational.
ῥητὴ ἄρα καὶ ἡ ΜΗ καὶ μήκει σύμμετρος τῇ ΜΛ, τουτέστι τῇ ΔΕ· ἀσύμμετρος ἄρα ἐστὶν ἡ ΔΜ τῇ ΜΗ μήκει.
And it is applied to the rational straight line ML; therefore MH is also rational and commensurable in length with ML, that is, with DE; therefore DM is incommensurable in length with MH.
καί εἰσι ῥηταί· αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ.
δεικτέον δή, ὅτι καὶ δευτέρα.
And they are rational; therefore DM, MH are rational straight lines commensurable in square only; therefore DH is a binomial straight line.
ἐπεὶ γὰρ τὰ ἀπὸ τῶν ΑΓ, ΓΒ μείζονά ἐστι τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, μεῖζον ἄρα καὶ τὸ ΔΛ τοῦ ΜΖ·
We must then prove that it is also a second binomial straight line.
ὥστε καὶ ἡ ΔΜ τῆς ΜΗ. καὶ ἐπεὶ σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΓ τῷ ἀπὸ τῆς ΓΒ, σύμμετρόν ἐστι καὶ τὸ ΔΘ τῷ ΚΛ·
For since the sum of the squares on AC, CB is greater than twice the rectangle contained by AC, CB, therefore DL is also greater than MZ; so that DM is also greater than MH.
ὥστε καὶ ἡ ΔΚ τῇ ΚΜ σύμμετρός ἐστιν.
And since the square on AC is commensurable with the square on CB, DΘ is also commensurable with KL; so that DK is also commensurable with KM.
καί ἐστι τὸ ὑπὸ τῶν ΔΚΜ ἴσον τῷ ἀπὸ τῆς ΜΝ· ἡ ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ.
And the rectangle contained by DK, KM is equal to the square on MN; therefore the square on DM is greater than the square on MH by the square on a straight line commensurable with DM.
καί ἐστιν ἡ ΜΗ σύμμετρος τῇ ΔΕ μήκει.
And MH is commensurable in length with DE.
ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ δευτέρα.
Therefore DH is a second binomial straight line.
§10.prop2.62τὸ ἀπὸ τῆς ἐκ δύο μέσων δευτέρας παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων τρίτην.
The square on a second bimedial straight line applied to a rational straight line produces as breadth a third binomial straight line.
ἔστω ἐκ δύο μέσων δευτέρα ἡ ΑΒ διῃρημένη εἰς τὰς μέσας κατὰ τὸ Γ, ὥστε τὸ μεῖζον τμῆμα εἶναι τὸ ΑΓ, ῥητὴ δέ τις ἔστω ἡ ΔΕ, καὶ παρὰ τὴν ΔΕ τῷ ἀπὸ τῆς ΑΒ ἴσον παραλληλόγραμμον παραβεβλήσθω τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ· λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ τρίτη.
Let AB be a second bimedial straight line divided into its medial terms at C, so that the greater segment is AC; and let DE be some 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 third binomial straight line.
Κατεσκευάσθω τὰ αὐτὰ τοῖς προδεδειγμένοις.
For let the same construction be made as in what was proved before.
καὶ ἐπεὶ ἐκ δύο μέσων δευτέρα ἐστὶν ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι μέσον περιέχουσαι· ὥστε καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ μέσον ἐστίν.
And since AB is a second bimedial straight line divided at C, therefore AC, CB are medial straight lines commensurable in square only containing a medial rectangle; so that the sum of the squares on AC, CB is also medial.
καί ἐστιν ἴσον τῷ ΔΛ· μέσον ἄρα καὶ τὸ ΔΛ. καὶ παράκειται παρὰ ῥητὴν τὴν ΔΕ·
And it is equal to DL; therefore DL is also medial.
ῥητὴ ἄρα ἐστὶ καὶ ἡ ΜΔ καὶ ἀσύμμετρος τῇ ΔΕ μήκει.
And it is applied to the rational straight line DE; therefore MD is also rational and incommensurable in length with DE.
διὰ τὰ αὐτὰ δὴ καὶ ἡ ΜΗ ῥητή ἐστι καὶ ἀσύμμετρος τῇ ΜΛ, τουτέστι τῇ ΔΕ, μήκει· ῥητὴ ἄρα ἐστὶν ἑκατέρα τῶν ΔΜ, ΜΗ καὶ ἀσύμμετρος τῇ ΔΕ μήκει.
For the same reason indeed, MH is also rational and incommensurable in length with ML, that is, with DE; therefore each of DM, MH is rational and incommensurable in length with DE.
καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΓ τῇ ΓΒ μήκει, ὡς δὲ ἡ ΑΓ πρὸς τὴν ΓΒ, οὕτως τὸ ἀπὸ τῆς ΑΓ πρὸς τὸ ὑπὸ τῶν ΑΓΒ, ἀσύμμετρον ἄρα καὶ τὸ ἀπὸ τῆς ΑΓ τῷ ὑπὸ τῶν ΑΓΒ. ὥστε καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ τῷ δὶς ὑπὸ τῶν ΑΓΒ ἀσύμμετρόν ἐστιν, τουτέστι τὸ ΔΛ τῷ ΜΖ·
And since AC is incommensurable in length with CB, and as AC is to CB, so is the square on AC to the rectangle contained by ACB, therefore the square on AC is also incommensurable with the rectangle contained by ACB.
ὥστε καὶ ἡ ΔΜ τῇ ΜΗ ἀσύμμετρός ἐστιν.
So that the sum of the squares on AC, CB is also incommensurable with twice the rectangle contained by ACB, that is, DL with MZ; so that DM is also incommensurable with MH.
καί εἰσι ῥηταί· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ.
δεικτέον, ὅτι καὶ τρίτη.
And they are rational; therefore DH is a binomial straight line. We must then prove that it is also a third binomial straight line.
ὁμοίως δὴ τοῖς προτέροις ἐπιλογιούμεθα, ὅτι μείζων ἐστὶν ἡ ΔΜ τῆς ΜΗ, καὶ σύμμετρος ἡ ΔΚ τῇ ΚΜ. καί ἐστι τὸ ὑπὸ τῶν ΔΚΜ ἴσον τῷ ἀπὸ τῆς ΜΝ·
Similarly indeed to the prior cases we will infer that DM is greater than MH, and DK is commensurable with KM.
ἡ ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ.
And the rectangle contained by DKM is equal to the square on MN; therefore the square on DM is greater than the square on MH by the square on a straight line commensurable with DM.
καὶ οὐδετέρα τῶν ΔΜ, ΜΗ σύμμετρός ἐστι τῇ ΔΕ μήκει.
And neither of DM, MH is commensurable in length with DE.
ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ τρίτη· ὅπερ ἔδει δεῖξαι.
Therefore DH is a third binomial straight line; which was to be proved.