§10.prop2.65τὸ ἀπὸ τῆς δύο μέσα δυναμένης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων ἕκτην.
The square on a straight line which produces two medial areas applied to a rational straight line produces as breadth a sixth binomial straight line.
ἔστω δύο μέσα δυναμένη ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, ῥητὴ δὲ ἔστω ἡ ΔΕ. καὶ παρὰ τὴν ΔΕ τῷ ἀπὸ τῆς ΑΒ ἴσον παραβεβλήσθω τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ·
Let AB be a straight line which produces two medial areas divided at C, 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 sixth binomial straight line.
λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶν ἕκτη. Κατεσκευάσθω γὰρ τὰ αὐτὰ τοῖς πρότερον.
For let the same construction be made as in what was proved before.
καὶ ἐπεὶ ἡ ΑΒ δύο μέσα δυναμένη ἐστὶ διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων μέσον καὶ τὸ ὑπʼ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τὸ ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων συγκείμενον τῷ ὑπʼ αὐτῶν· ὥστε κατὰ τὰ προδεδειγμένα μέσον ἐστὶν ἑκάτερον τῶν ΔΛ, ΜΖ. καὶ παρὰ ῥητὴν τὴν ΔΕ παράκειται·
And since AB is a straight line which produces two medial areas divided at C, therefore AC, CB are incommensurable in square, making the sum of the squares on them medial, and the rectangle contained by them medial, and moreover the sum of the squares on them incommensurable with the rectangle contained by them; so that, according to what was proved before, each of DL, MZ is medial.
ῥητὴ ἄρα ἐστὶν ἑκατέρα τῶν ΔΜ, ΜΗ καὶ ἀσύμμετρος τῇ ΔΕ μήκει.
And they are applied to the rational straight line DE; therefore each of DM, MH is rational and incommensurable in length with DE.
καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, ἀσύμμετρον ἄρα ἐστὶ τὸ ΔΛ τῷ ΜΖ. ἀσύμμετρος ἄρα καὶ ἡ ΔΜ τῇ ΜΗ·
And since the sum of the squares on AC, CB is incommensurable with twice the rectangle contained by AC, CB, therefore DL is also incommensurable with MZ.
αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ.
λέγω δή, ὅτι καὶ ἕκτη.
Therefore DM is also incommensurable with MH; therefore DM, MH are rational straight lines commensurable in square only; therefore DH is a binomial straight line.
ὁμοίως δὴ πάλιν δείξομεν, ὅτι τὸ ὑπὸ τῶν ΔΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΜΝ, καὶ ὅτι ἡ ΔΚ τῇ ΚΜ μήκει ἐστὶν ἀσύμμετρος· καὶ διὰ τὰ αὐτὰ δὴ ἡ ΔΜ τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει.
I say next that it is also a sixth. For we will similarly prove again that the rectangle contained by DKM is equal to the square on MN, and that DK is incommensurable in length with KM; and for the same reasons indeed the square on DM is greater than the square on MH by the square on a straight line incommensurable in length with DM.
καὶ οὐδετέρα τῶν ΔΜ, ΜΗ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΔΕ μήκει.
And neither of DM, MH is commensurable in length with the set out rational straight line DE.
ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶν ἕκτη· ὅπερ ἔδει δεῖξαι.
Therefore DH is a sixth binomial straight line; which was to be proved.
§10.prop2.66ἡ τῇ ἐκ δύο ὀνομάτων μήκει σύμμετρος καὶ αὐτὴ ἐκ δύο ὀνομάτων ἐστὶ καὶ τῇ τάξει ἡ αὐτή.
A straight line commensurable in length with a binomial straight line is itself also binomial and the same in order.
ἔστω ἐκ δύο ὀνομάτων ἡ ΑΒ, καὶ τῇ ΑΒ μήκει σύμμετρος ἔστω ἡ ΓΔ· λέγω, ὅτι ἡ ΓΔ ἐκ δύο ὀνομάτων ἐστὶ καὶ τῇ τάξει ἡ αὐτὴ τῇ ΑΒ.
ἐπεὶ γὰρ ἐκ δύο ὀνομάτων ἐστὶν ἡ ΑΒ, διῃρήσθω εἰς τὰ ὀνόματα κατὰ τὸ Ε, καὶ ἔστω μεῖζον ὄνομα τὸ ΑΕ· αἱ ΑΕ, ΕΒ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι.
Let AB be a binomial straight line, and let CD be commensurable in length with AB; I say that CD is binomial and the same in order with AB. For since AB is binomial, let it be divided into its terms at E, and let AE be the greater term; therefore AE, EB are rational straight lines commensurable in square only.
γεγονέτω ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΑΕ πρὸς τὴν ΓΖ· καὶ λοιπὴ ἄρα ἡ ΕΒ πρὸς λοιπὴν τὴν ΖΔ ἐστιν, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ. σύμμετρος δὲ ἡ ΑΒ τῇ ΓΔ μήκει. σύμμετρος ἄρα ἐστὶ καὶ ἡ μὲν ΑΕ τῇ ΓΖ, ἡ δὲ ΕΒ τῇ ΖΔ. καί εἰσι ῥηταὶ αἱ ΑΕ, ΕΒ· ῥηταὶ ἄρα εἰσὶ καὶ αἱ ΓΖ, ΖΔ. καὶ ἐστιν ὡς ἡ ΑΕ πρὸς ΓΖ, ἡ ΕΒ πρὸς ΖΔ. ἐναλλὰξ ἄρα ἐστὶν ὡς ἡ ΑΕ πρὸς ΕΒ, ἡ ΓΖ πρὸς ΖΔ. αἱ δὲ ΑΕ, ΕΒ δυνάμει μόνον σύμμετροι· καὶ αἱ ΓΖ, ΖΔ ἄρα δυνάμει μόνον εἰσὶ σύμμετροι.
Let it be made: as AB is to CD, so AE to CZ; therefore the remainder EB is also to the remainder ZD as AB is to CD. But AB is commensurable in length with CD; therefore AE is also commensurable with CZ, and EB with ZD. And AE, EB are rational; therefore CZ, ZD are also rational. And as AE is to CZ, so is EB to ZD; therefore alternately, as AE is to EB, so is CZ to ZD. But AE, EB are commensurable in square only; therefore CZ, ZD are also commensurable in square only.
καί εἰσι ῥηταί· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΓΔ.
λέγω δή, ὅτι τῇ τάξει ἐστὶν ἡ αὐτὴ τῇ ΑΒ.
ἡ γὰρ ΑΕ τῆς ΕΒ μεῖζον δύναται ἤτοι τῷ ἀπὸ συμμέτρου ἑαυτῇ ἢ τῷ ἀπὸ ἀσυμμέτρου. εἰ μὲν οὖν ἡ ΑΕ τῆς ΕΒ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ ἡ ΓΖ τῆς ΖΔ μεῖζον δυνήσεται τῷ ἀπὸ συμμέτρου ἑαυτῇ.
And they are rational; therefore CD is a binomial straight line. I say next that it is also the same in order with AB. For the square on AE is greater than the square on EB either by the square on a straight line commensurable in length with AE, or by the square on a straight line incommensurable with it. If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, then the square on CZ will also be greater than the square on ZD by the square on a straight line commensurable with CZ.
καὶ εἰ μὲν σύμμετρός ἐστιν ἡ ΑΕ τῇ ἐκκειμένῃ ῥητῇ, καὶ ἡ ΓΖ σύμμετρος αὐτῇ ἔσται, καὶ διὰ τοῦτο ἑκατέρα τῶν ΑΒ, ΓΔ ἐκ δύο ὀνομάτων ἐστὶ πρώτη, τουτέστι τῇ τάξει ἡ αὐτή.
And if AE is commensurable with the set out rational straight line, CZ will also be commensurable with it, and for this reason each of AB, CD is a first binomial, that is, the same in order.
εἰ δὲ ἡ ΕΒ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ, καὶ ἡ ΖΔ σύμμετρός ἐστιν αὐτῇ, καὶ διὰ τοῦτο πάλιν τῇ τάξει ἡ αὐτὴ ἔσται τῇ ΑΒ· ἑκατέρα γὰρ αὐτῶν ἔσται ἐκ δύο ὀνομάτων δευτέρα.
But if EB is commensurable with the set out rational straight line, ZD will also be commensurable with it, and for this reason again it will be the same in order with AB; for each of them will be a second binomial.
εἰ δὲ οὐδετέρα τῶν ΑΕ, ΕΒ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ, οὐδετέρα τῶν ΓΖ, ΖΔ σύμμετρος αὐτῇ ἔσται, καί ἐστιν ἑκατέρα τρίτη.
But if neither of AE, EB is commensurable with the set out rational straight line, neither of CZ, ZD will be commensurable with it, and each is a third binomial.
εἰ δὲ ἡ ΑΕ τῆς ΕΒ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ ἡ ΓΖ τῆς ΖΔ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ.
But if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, then the square on CZ is also greater than the square on ZD by the square on a straight line incommensurable with CZ.
καὶ εἰ μὲν ἡ ΑΕ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ, καὶ ἡ ΓΖ σύμμετρός ἐστιν αὐτῇ, καί ἐστιν ἑκατέρα τετάρτη.
And if AE is commensurable with the set out rational straight line, CZ is also commensurable with it, and each is a fourth binomial.
εἰ δὲ ἡ ΕΒ, καὶ ἡ ΖΔ, καὶ ἔσται ἑκατέρα πέμπτη.
But if EB is, then ZD is also, and each will be a fifth binomial.
εἰ δὲ οὐδετέρα τῶν ΑΕ, ΕΒ, καὶ τῶν ΓΖ, ΖΔ οὐδετέρα σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ, καὶ ἔσται ἑκατέρα ἕκτη.
But if neither of AE, EB is, then neither of CZ, ZD is commensurable with the set out rational straight line, and each will be a sixth binomial.
ὥστε ἡ τῇ ἐκ δύο ὀνομάτων μήκει σύμμετρος ἐκ δύο ὀνομάτων ἐστὶ καὶ τῇ τάξει ἡ αὐτή· ὅπερ ἔδει δεῖξαι.
Therefore a straight line commensurable in length with a binomial straight line is binomial and the same in order; which was to be proved.