§10.prop2.53#2ἔστω οὖν τῷ ἀπὸ ΖΗ ἴσα τὰ ἀπὸ τῶν ΗΘ, Κ·
Let therefore the squares on HT, K be equal to the square on ZH.
ἀναστρέψαντι ἄρα ὡς ὁ ΑΒ πρὸς ΒΓ, οὕτως τὸ ἀπὸ ΖΗ πρὸς τὸ ἀπὸ τῆς Κ. ὁ δὲ ΑΒ πρὸς τὸν ΒΓ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν·
Therefore, by conversion, as AB is to BC, so is the square on ZH to the square on K.
ὥστε οὐδὲ τὸ ἀπὸ ΖΗ πρὸς τὸ ἀπὸ τῆς Κ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν.
But AB has not to BC the ratio which a square number has to a square number; so that neither has the square on ZH to the square on K the ratio which a square number has to a square number.
ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΗ τῇ Κ μήκει· ἡ ΖΗ ἄρα τῆς ΗΘ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ.
Therefore ZH is incommensurable in length with K; therefore ZH is greater in square than HT by the square on a straight line incommensurable in length with itself.
καί εἰσιν αἱ ΖΗ, ΗΘ ῥηταὶ δυνάμει μόνον σύμμετροι, καὶ οὐδετέρα αὐτῶν σύμμετρός ἐστι μήκει τῇ ἐκκειμένῃ ῥητῇ τῇ Ε.
ἡ ΖΘ ἄρα ἐκ δύο ὀνομάτων ἐστὶν ἕκτη· ὅπερ ἔδει δεῖξαι.
And ZH, HT are rational straight lines commensurable in square only, and neither of them is commensurable in length with the set-out rational straight line E. Therefore ZT is a sixth binomial straight line; which was to be proved.
λῆμμα
ἔστω δύο τετράγωνα τὰ ΑΒ, ΒΓ καὶ κείσθωσαν ὥστε ἐπʼ εὐθείας εἶναι τὴν ΔΒ τῇ ΒΕ· ἐπʼ εὐθείας ἄρα ἐστὶ καὶ ἡ ΖΒ τῇ ΒΗ. καὶ συμπεπληρώσθω τὸ ΑΓ παραλληλόγραμμον·
Lemma Let AB, BC be two squares, and let them be placed so that DB is in a straight line with BE; therefore ZB is also in a straight line with BH.
λέγω, ὅτι τετράγωνόν ἐστι τὸ ΑΓ, καὶ ὅτι τῶν ΑΒ, ΒΓ μέσον ἀνάλογόν ἐστι τὸ ΔΗ, καὶ ἔτι τῶν ΑΓ, ΓΒ μέσον ἀνάλογόν ἐστι τὸ ΔΓ.
ἐπεὶ γὰρ ἴση ἐστὶν ἡ μὲν ΔΒ τῇ ΒΖ, ἡ δὲ ΒΕ τῇ ΒΗ, ὅλη ἄρα ἡ ΔΕ ὅλῃ τῇ ΖΗ ἐστιν ἴση.
And let the parallelogram AC be completed; I say that AC is a square, and that DH is a mean proportional between AB, BC, and further that DC is a mean proportional between AC, CB. For since DB is equal to BZ, and BE to BH, therefore the whole DE is equal to the whole ZH.
ἀλλʼ ἡ μὲν ΔΕ ἑκατέρᾳ τῶν ΑΘ, ΚΓ ἐστιν ἴση, ἡ δὲ ΖΗ ἑκατέρᾳ τῶν ΑΚ, ΘΓ ἐστιν ἴση· καὶ ἑκατέρα ἄρα τῶν ΑΘ, ΚΓ ἑκατέρᾳ τῶν ΑΚ, ΘΓ ἐστιν ἴση.
But DE is equal to each of AT, KC, and ZH is equal to each of AK, TC; therefore each of AT, KC is also equal to each of AK, TC.
ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΓ παραλληλόγραμμον· ἔστι δὲ καὶ ὀρθογώνιον·
Therefore the parallelogram AC is equilateral; and it is also right-angled; therefore AC is a square.
τετράγωνον ἄρα ἐστὶ τὸ ΑΓ.
καὶ ἐπεί ἐστιν ὡς ἡ ΖΒ πρὸς τὴν ΒΗ, οὕτως ἡ ΔΒ πρὸς τὴν ΒΕ, ἀλλʼ ὡς μὲν ἡ ΖΒ πρὸς τὴν ΒΗ, οὕτως τὸ ΑΒ πρὸς τὸ ΔΗ, ὡς δὲ ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως τὸ ΔΗ πρὸς τὸ ΒΓ, καὶ ὡς ἄρα τὸ ΑΒ πρὸς τὸ ΔΗ, οὕτως τὸ ΔΗ πρὸς τὸ ΒΓ. τῶν ΑΒ, ΒΓ ἄρα μέσον ἀνάλογόν ἐστι τὸ ΔΗ.
λέγω δή, ὅτι καὶ τῶν ΑΓ, ΓΒ μέσον ἀνάλογόν τὸ ΔΓ.
ἐπεὶ γάρ ἐστιν ὡς ἡ ΑΔ πρὸς τὴν ΔΚ, οὕτως ἡ ΚΗ πρὸς τὴν ΗΓ·
And since as ZB is to BH, so is DB to BE, but as ZB is to BH, so is AB to DH, and as DB is to BE, so is DH to BC, therefore also as AB is to DH, so is DH to BC. Therefore DH is a mean proportional between AB, BC. I say then that DC is also a mean proportional between AC, CB.
ἴση γάρ ἑκατέρα ἑκατέρᾳ·
For since as AD is to DK, so is KH to HC; for each is equal to each; and, by addition, as AK is to KD, so is KC to CH, but as AK is to KD, so is AC to CD, and as KC is to CH, so is DC to CB, therefore also as AC is to DC, so is DC to BC.
καὶ συνθέντι ὡς ἡ ΑΚ πρὸς ΚΔ, οὕτως ἡ ΚΓ πρὸς ΓΗ, ἀλλʼ ὡς μὲν ἡ ΑΚ πρὸς ΚΔ, οὕτως τὸ ΑΓ πρὸς τὸ ΓΔ, ὡς δὲ ἡ ΚΓ πρὸς ΓΗ, οὕτως τὸ ΔΓ πρὸς ΓΒ, καὶ ὡς ἄρα τὸ ΑΓ πρὸς ΔΓ, οὕτως τὸ ΔΓ πρὸς τὸ ΒΓ. τῶν ΑΓ, ΓΒ ἄρα μέσον ἀνάλογόν ἐστι τὸ ΔΓ· ἃ προέκειτο δεῖξαι.
Therefore DC is a mean proportional between AC, CB; which it was proposed to prove.