§10.prop2.54ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων πρώτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη ἐκ δύο ὀνομάτων.
If an area be contained by a rational straight line and a first binomial, the side of the square equal to the area is the irrational straight line called binomial.
χωρίον γὰρ τὸ ΑΓ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΒ καὶ τῆς ἐκ δύο ὀνομάτων πρώτης τῆς ΑΔ· λέγω, ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη ἐκ δύο ὀνομάτων.
For let the area AC be contained by the rational straight line AB and the first binomial AD; I say that the side of the square equal to the area AC is the irrational straight line called binomial.
ἐπεὶ γὰρ ἐκ δύο ὀνομάτων ἐστὶ πρώτη ἡ ΑΔ, διῃρήσθω εἰς τὰ ὀνόματα κατὰ τὸ Ε, καὶ ἔστω τὸ μεῖζον ὄνομα τὸ ΑΕ. φανερὸν δή, ὅτι αἱ ΑΕ, ΕΔ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ ΑΕ τῆς ΕΔ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ ἡ ΑΕ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΑΒ μήκει.
For since AD is a first binomial straight line, let it be divided into its terms at E, and let AE be the greater term. It is then manifest that AE, ED are rational straight lines commensurable in square only, and AE is greater in square than ED by the square on a straight line commensurable in length with itself, and AE is commensurable in length with the set-out rational straight line AB.
τετμήσθω δὴ ἡ ΕΔ δίχα κατὰ τὸ Ζ σημεῖον.
Let then ED be bisected at the point Z.
καὶ ἐπεὶ ἡ ΑΕ τῆς ΕΔ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, ἐὰν ἄρα τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ἐλάσσονος, τουτέστι τῷ ἀπὸ τῆς ΕΖ, ἴσον παρὰ τὴν μείζονα τὴν ΑΕ παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς σύμμετρα αὐτὴν διαιρεῖ.
And since AE is greater in square than ED by the square on a straight line commensurable with itself, therefore, if an area equal to the fourth part of the square on the less, that is, to the square on EZ, be applied to the greater AE, falling short by a square figure, it divides it into parts commensurable with itself.
παραβεβλήσθω οὖν παρὰ τὴν ΑΕ τῷ ἀπὸ τῆς ΕΖ ἴσον τὸ ὑπὸ ΑΗ, ΗΕ· σύμμετρος ἄρα ἐστὶν ἡ ΑΗ τῇ ΕΗ μήκει.
Let then there be applied to AE the rectangle contained by AH, HE equal to the square on EZ; therefore AH is commensurable in length with EH.
καὶ ἤχθωσαν ἀπὸ τῶν Η, Ε, Ζ ὁποτέρᾳ τῶν ΑΒ, ΓΔ παράλληλοι αἱ ΗΘ, ΕΚ, ΖΛ· καὶ τῷ μὲν ΑΘ παραλληλογράμμῳ ἴσον τετράγωνον συνεστάτω τὸ ΣΝ, τῷ δὲ ΗΚ ἴσον τὸ ΝΠ, καὶ κείσθω ὥστε ἐπʼ εὐθείας εἶναι τὴν ΜΝ τῇ ΝΞ· ἐπʼ εὐθείας ἄρα ἐστὶ καὶ ἡ ΡΝ τῇ ΝΟ. καὶ συμπεπληρώσθω τὸ ΣΠ παραλληλόγραμμον· τετράγωνον ἄρα ἐστὶ τὸ ΣΠ. καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΗ, ΗΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΖ, ἔστιν ἄρα ὡς ἡ ΑΗ πρὸς ΕΖ, οὕτως ἡ ΖΕ πρὸς ΕΗ· καὶ ὡς ἄρα τὸ ΑΘ πρὸς ΕΛ, τὸ ΕΛ πρὸς ΚΗ·
And let there be drawn from H, E, Z, parallel to either of AB, CD, the straight lines HT, EK, ZL; and let there be constructed, equal to the parallelogram AT, the square SN, and, equal to HK, the square NP, and let them be placed so that MN is in a straight line with NX; therefore RN is also in a straight line with NO. And let the parallelogram SP be completed; therefore SP is a square. And since the rectangle contained by AH, HE is equal to the square on EZ, therefore as AH is to EZ, so is ZE to EH; and therefore as AT is to EL, so is EL to KH; therefore EL is a mean proportional between AT, HK.
τῶν ΑΘ, ΗΚ ἄρα μέσον ἀνάλογόν ἐστι τὸ ΕΛ. ἀλλὰ τὸ μὲν ΑΘ ἴσον ἐστὶ τῷ ΣΝ, τὸ δὲ ΗΚ ἴσον τῷ ΝΠ·
But AT is equal to SN, and HK to NP; therefore EL is a mean proportional between SN, NP.
τῶν ΣΝ, ΝΠ ἄρα μέσον ἀνάλογόν ἐστι τὸ ΕΛ. ἔστι δὲ τῶν αὐτῶν τῶν ΣΝ, ΝΠ μέσον ἀνάλογον καὶ τὸ ΜΡ· ἴσον ἄρα ἐστὶ τὸ ΕΛ τῷ ΜΡ· ὥστε καὶ τῷ ΟΞ ἴσον ἐστίν.
But MP is also a mean proportional between the same SN, NP; therefore EL is equal to MP; so that it is also equal to OX.
ἔστι δὲ καὶ τὰ ΑΘ, ΗΚ τοῖς ΣΝ, ΝΠ ἴσα· ὅλον ἄρα τὸ ΑΓ ἴσον ἐστὶν ὅλῳ τῷ ΣΠ, τουτέστι τῷ ἀπὸ τῆς ΜΞ τετραγώνῳ·
And AT, HK are also equal to SN, NP; therefore the whole AC is equal to the whole SP, that is, to the square on MX; therefore MX is the side of the square equal to AC.
τὸ ΑΓ ἄρα δύναται ἡ ΜΞ.
λέγω, ὅτι ἡ ΜΞ ἐκ δύο ὀνομάτων ἐστίν.
I say that MX is a binomial straight line.
ἐπεὶ γὰρ σύμμετρός ἐστιν ἡ ΑΗ τῇ ΗΕ, σύμμετρός ἐστι καὶ ἡ ΑΕ ἑκατέρᾳ τῶν ΑΗ, ΗΕ. ὑπόκειται δὲ καὶ ἡ ΑΕ τῇ ΑΒ σύμμετρος· καὶ αἱ ΑΗ, ΗΕ ἄρα τῇ ΑΒ σύμμετροί εἰσιν.
For since AH is commensurable with HE, AE is also commensurable with each of AH, HE. But AE is also assumed to be commensurable with AB; therefore AH, HE are also commensurable with AB.
καί ἐστι ῥητὴ ἡ ΑΒ· ῥητὴ ἄρα ἐστὶ καὶ ἑκατέρα τῶν ΑΗ, ΗΕ· ῥητὸν ἄρα ἐστὶν ἑκάτερον τῶν ΑΘ, ΗΚ, καί ἐστι σύμμετρον τὸ ΑΘ τῷ ΗΚ. ἀλλὰ τὸ μὲν ΑΘ τῷ ΣΝ ἴσον ἐστίν, τὸ δὲ ΗΚ τῷ ΝΠ· καὶ τὰ ΣΝ, ΝΠ ἄρα, τουτέστι τὰ ἀπὸ τῶν ΜΝ, ΝΞ, ῥητά ἐστι καὶ σύμμετρα.
And AB is rational; therefore each of AH, HE is also rational; therefore each of AT, HK is rational, and AT is commensurable with HK. But AT is equal to SN, and HK to NP; therefore SN, NP also, that is, the squares on MN, NX, are rational and commensurable.
καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΕ τῇ ΕΔ μήκει, ἀλλʼ ἡ μὲν ΑΕ τῇ ΑΗ ἐστι σύμμετρος, ἡ δὲ ΔΕ τῇ ΕΖ σύμμετρος, ἀσύμμετρος ἄρα καὶ ἡ ΑΗ τῇ ΕΖ· ὥστε καὶ τὸ ΑΘ τῷ ΕΛ ἀσύμμετρόν ἐστιν.
And since AE is incommensurable in length with ED, but AE is commensurable with AH, and DE is commensurable with EZ, therefore AH is also incommensurable with EZ; so that AT is also incommensurable with EL.
ἀλλὰ τὸ μὲν ΑΘ τῷ ΣΝ ἐστιν ἴσον, τὸ δὲ ΕΛ τῷ ΜΡ· καὶ τὸ ΣΝ ἄρα τῷ ΜΡ ἀσύμμετρόν ἐστιν.
But AT is equal to SN, and EL to MP; therefore SN is also incommensurable with MP.
ἀλλʼ ὡς τὸ ΣΝ πρὸς ΜΡ, ἡ ΟΝ πρὸς τὴν ΝΡ·
But as SN is to MP, so is ON to NP; therefore ON is incommensurable with NP.
ἀσύμμετρος ἄρα ἐστὶν ἡ ΟΝ τῇ ΝΡ. ἴση δὲ ἡ μὲν ΟΝ τῇ ΜΝ, ἡ δὲ ΝΡ τῇ ΝΞ·
But ON is equal to MN, and NP to NX; therefore MN is incommensurable with NX.
ἀσύμμετρος ἄρα ἐστὶν ἡ ΜΝ τῇ ΝΞ. καί ἐστι τὸ ἀπὸ τῆς ΜΝ σύμμετρον τῷ ἀπὸ τῆς ΝΞ, καὶ ῥητὸν ἑκάτερον· αἱ ΜΝ, ΝΞ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι.
And the square on MN is commensurable with the square on NX, and each is rational; therefore MN, NX are rational straight lines commensurable in square only.
ἡ ΜΞ ἄρα ἐκ δύο ὀνομάτων ἐστὶ καὶ δύναται τὸ ΑΓ· ὅπερ ἔδει δεῖξαι.
Therefore MX is a binomial straight line, and is the side of the square equal to AC; which was to be proved.