§10.prop2.55ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων δευτέρας, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη ἐκ δύο μέσων πρώτη.
If an area be contained by a rational straight line and a second binomial, the side of the square equal to the area is the irrational straight line called first bimedial.
περιεχέσθω γὰρ χωρίον τὸ ΑΒΓΔ ὑπὸ ῥητῆς τῆς ΑΒ καὶ τῆς ἐκ δύο ὀνομάτων δευτέρας τῆς ΑΔ· λέγω, ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἐκ δύο μέσων πρώτη ἐστίν.
For let the area ABCD be contained by the rational straight line AB and the second binomial AD; I say that the side of the square equal to the area AC is a first bimedial straight line.
ἐπεὶ γὰρ ἐκ δύο ὀνομάτων δευτέρα ἐστὶν ἡ ΑΔ, διῃρήσθω εἰς τὰ ὀνόματα κατὰ τὸ Ε, ὥστε τὸ μεῖζον ὄνομα εἶναι τὸ ΑΕ· αἱ ΑΕ, ΕΔ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ ΑΕ τῆς ΕΔ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ τὸ ἔλαττον ὄνομα ἡ ΕΔ σύμμετρόν ἐστι τῇ ΑΒ μήκει.
For since AD is a second binomial, let it be divided into its terms at E, so that the greater term is AE; therefore 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 the lesser term ED is commensurable in length with AB.
τετμήσθω ἡ ΕΔ δίχα κατὰ τὸ Ζ, καὶ τῷ ἀπὸ τῆς ΕΖ ἴσον παρὰ τὴν ΑΕ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ τὸ ὑπὸ τῶν ΑΗΕ· σύμμετρος ἄρα ἡ ΑΗ τῇ ΗΕ μήκει.
Let ED be bisected at Z, and let there be applied to AE, falling short by a square figure, a rectangle equal to the square on EZ, namely that contained by AH, HE; therefore AH is commensurable in length with HE.
καὶ διὰ τῶν Η, Ε, Ζ παράλληλοι ἤχθωσαν ταῖς ΑΒ, ΓΔ αἱ ΗΘ, ΕΚ, ΖΛ, καὶ τῷ μὲν ΑΘ παραλληλογράμμῳ ἴσον τετράγωνον συνεστάτω τὸ ΣΝ, τῷ δὲ ΗΚ ἴσον τετράγωνον τὸ ΝΠ, καὶ κείσθω ὥστε ἐπʼ εὐθείας εἶναι τὴν ΜΝ τῇ ΝΞ· ἐπʼ εὐθείας ἄρα καὶ ἡ ΡΝ τῇ ΝΟ. καὶ συμπεπληρώσθω τὸ ΣΠ τετράγωνον·
And through H, E, Z let there be drawn parallel to 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.
φανερὸν δὴ ἐκ τοῦ προδεδειγμένου, ὅτι τὸ ΜΡ μέσον ἀνάλογόν ἐστι τῶν ΣΝ, ΝΠ, καὶ ἴσον τῷ ΕΛ, καὶ ὅτι τὸ ΑΓ χωρίον δύναται ἡ ΜΞ. δεικτέον δή, ὅτι ἡ ΜΞ ἐκ δύο μέσων ἐστὶ πρώτη.
And let the square SP be completed; it is then manifest from what was proved before that MP is a mean proportional between SN, NP, and is equal to EL, and that the side of the square equal to the area AC is MX. We must then prove that MX is a first bimedial straight line.
ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΕ τῇ ΕΔ μήκει, σύμμετρος δὲ ἡ ΕΔ τῇ ΑΒ, ἀσύμμετρος ἄρα ἡ ΑΕ τῇ ΑΒ. καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΑΗ τῇ ΕΗ, σύμμετρός ἐστι καὶ ἡ ΑΕ ἑκατέρᾳ τῶν ΑΗ, ΗΕ. ἀλλὰ ἡ ΑΕ ἀσύμμετρος τῇ ΑΒ μήκει· καὶ αἱ ΑΗ, ΗΕ ἄρα ἀσύμμετροί εἰσι τῇ ΑΒ. αἱ ΒΑ, ΑΗ, ΗΕ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ὥστε μέσον ἐστὶν ἑκάτερον τῶν ΑΘ, ΗΚ. ὥστε καὶ ἑκάτερον τῶν ΣΝ, ΝΠ μέσον ἐστίν.
Since AE is incommensurable in length with ED, and ED is commensurable with AB, therefore AE is incommensurable with AB. And since AH is commensurable with EH, AE is also commensurable with each of AH, HE. But AE is incommensurable in length with AB; therefore AH, HE are also incommensurable with AB. Therefore BA, AH, HE are rational straight lines commensurable in square only; so that each of AT, HK is medial. So that each of SN, NP is also medial.
καὶ αἱ ΜΝ, ΝΞ ἄρα μέσαι εἰσίν.
Therefore MN, NX are also medial.
καὶ ἐπεὶ σύμμετρος ἡ ΑΗ τῇ ΗΕ μήκει, σύμμετρόν ἐστι καὶ τὸ ΑΘ τῷ ΗΚ, τουτέστι τὸ ΣΝ τῷ ΝΠ, τουτέστι τὸ ἀπὸ τῆς ΜΝ τῷ ἀπὸ τῆς ΝΞ.
And since AH is commensurable in length with HE, AT is also commensurable with HK, that is, SN with NP, that is, the square on MN with the square on NX.
καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΕ τῇ ΕΔ μήκει, ἀλλʼ ἡ μὲν ΑΕ σύμμετρός ἐστι τῇ ΑΗ, ἡ δὲ ΕΔ τῇ ΕΖ σύμμετρος, ἀσύμμετρος ἄρα ἡ ΑΗ τῇ ΕΖ· ὥστε καὶ τὸ ΑΘ τῷ ΕΛ ἀσύμμετρόν ἐστιν, τουτέστι τὸ ΣΝ τῷ ΜΡ, τουτέστιν ἡ ΟΝ τῇ ΝΡ, τουτέστιν ἡ ΜΝ τῇ ΝΞ ἀσύμμετρός ἐστι μήκει.
And since AE is incommensurable in length with ED, while AE is commensurable with AH, and ED is commensurable with EZ, therefore AH is also incommensurable with EZ; so that AT is also incommensurable with EL, that is, SN with MP, that is, ON with NP, that is, MN is incommensurable in length with NX.
ἐδείχθησαν δὲ αἱ ΜΝ, ΝΞ καὶ μέσαι οὖσαι καὶ δυνάμει σύμμετροι· αἱ ΜΝ, ΝΞ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι.
But MN, NX were also proved to be medial and commensurable in square; therefore MN, NX are medial straight lines commensurable in square only.
λέγω δή, ὅτι καὶ ῥητὸν περιέχουσιν.
I say then that they also contain a rational area.
ἐπεὶ γὰρ ἡ ΔΕ ὑπόκειται ἑκατέρᾳ τῶν ΑΒ, ΕΖ σύμμετρος, σύμμετρος ἄρα καὶ ἡ ΕΖ τῇ ΕΚ. καὶ ῥητὴ ἑκατέρα αὐτῶν· ῥητὸν ἄρα τὸ ΕΛ, τουτέστι τὸ ΜΡ·
For since DE is assumed to be commensurable with each of AB, EZ, therefore EZ is also commensurable with EK. And each of them is rational; therefore EL, that is, MP, is rational.
τὸ δὲ ΜΡ ἐστι τὸ ὑπὸ τῶν ΜΝΞ. ἐὰν δὲ δύο μέσαι δυνάμει μόνον σύμμετροι συντεθῶσι ῥητὸν περιέχουσαι, ἡ ὅλη ἄλογός ἐστιν, καλεῖται δὲ ἐκ δύο μέσων πρώτη.
And MP is the rectangle contained by MN, NX. But if two medial straight lines commensurable in square only and containing a rational area be added together, the whole is irrational, and is called a first bimedial straight line.
ἡ ἄρα ΜΞ ἐκ δύο μέσων ἐστὶ πρώτη· ὅπερ ἔδει δεῖξαι.
Therefore MX is a first bimedial straight line; which was to be proved.