§10.prop1.38ἐὰν δύο μέσαι δυνάμει μόνον σύμμετροι συντεθῶσι μέσον περιέχουσαι, ἡ ὅλη ἄλογός ἐστιν, καλείσθω δὲ ἐκ δύο μέσων δευτέρα.
If two medial straight lines commensurable in square only containing a medial area be added together, the whole is irrational, and let it be called the second bimedial.
Συγκείσθωσαν γὰρ δύο μέσαι δυνάμει μόνον σύμμετροι αἱ ΑΒ, ΒΓ μέσον περιέχουσαι· λέγω, ὅτι ἄλογός ἐστιν ἡ ΑΓ.
Ἐκκείσθω γὰρ ῥητὴ ἡ ΔΕ, καὶ τῷ ἀπὸ τῆς ΑΓ ἴσον παρὰ τὴν ΔΕ παραβεβλήσθω τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ. καὶ ἐπεὶ τὸ ἀπὸ τῆς ΑΓ ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν ΑΒ, ΒΓ καὶ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, παραβεβλήσθω δὴ τοῖς ἀπὸ τῶν ΑΒ, ΒΓ παρὰ τὴν ΔΕ ἴσον τὸ ΕΘ·
For let two medial straight lines commensurable in square only AB, BC containing a medial area be added together; I say that AC is irrational. For let the rational straight line DE be set out, and let the rectangle DZ equal to the square on AC be applied to DE, producing DH as width. And since the square on AC is equal to the sum of the squares on AB, BC and twice the rectangle contained by AB, BC, let the rectangle ETh equal to the sum of the squares on AB, BC be applied to DE.
λοιπὸν ἄρα τὸ ΘΖ ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. καὶ ἐπεὶ μέση ἐστὶν ἑκατέρα τῶν ΑΒ, ΒΓ, μέσα ἄρα ἐστὶ καὶ τὰ ἀπὸ τῶν ΑΒ, ΒΓ. μέσον δὲ ὑπόκειται καὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. καί ἐστι τοῖς μὲν ἀπὸ τῶν ΑΒ, ΒΓ ἴσον τὸ ΕΘ, τῷ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ἴσον τὸ ΖΘ·
Therefore the remainder, the rectangle ThZ, is equal to twice the rectangle contained by AB, BC. And since each of AB, BC is medial, therefore the squares on AB, BC are also medial. But twice the rectangle contained by AB, BC is also by hypothesis medial. And ETh is equal to the sum of the squares on AB, BC, and ZTh is equal to twice the rectangle contained by AB, BC.
μέσον ἄρα ἑκάτερον τῶν ΕΘ, ΘΖ. καὶ παρὰ ῥητὴν τὴν ΔΕ παράκειται· ῥητὴ ἄρα ἐστὶν ἑκατέρα τῶν ΔΘ, ΘΗ καὶ ἀσύμμετρος τῇ ΔΕ μήκει.
Therefore each of ETh, ThZ is medial. And they are applied to the rational straight line DE; therefore each of DTh, ThH is a rational straight line and incommensurable in length with DE.
ἐπεὶ οὖν ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει, καί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ὑπὸ τῶν ΑΒ, ΒΓ, ἀσύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΒ τῷ ὑπὸ τῶν ΑΒ, ΒΓ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΒ σύμμετρόν ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ τετραγώνων, τῷ δὲ ὑπὸ τῶν ΑΒ, ΒΓ σύμμετρόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. ἀσύμμετρον ἄρα ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΑΒ, ΒΓ ἴσον ἐστὶ τὸ ΕΘ, τῷ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ἴσον ἐστὶ τὸ ΘΖ. ἀσύμμετρον ἄρα ἐστὶ τὸ ΕΘ τῷ ΘΖ· ὥστε καὶ ἡ ΔΘ τῇ ΘΗ ἐστιν ἀσύμμετρος μήκει.
Since then AB is incommensurable in length with BC, and as AB is to BC, so is the square on AB to the rectangle contained by AB, BC, therefore the square on AB is incommensurable with the rectangle contained by AB, BC. But the square on AB is commensurable with the sum of the squares on AB, BC, and the rectangle contained by AB, BC is commensurable with twice the rectangle contained by AB, BC. Therefore the sum of the squares on AB, BC is incommensurable with twice the rectangle contained by AB, BC. But ETh is equal to the sum of the squares on AB, BC, and ThZ is equal to twice the rectangle contained by AB, BC. Therefore ETh is incommensurable with ThZ; so that DTh is also incommensurable in length with ThH.
αἱ ΔΘ, ΘΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι.
Therefore DTh, ThH are rational straight lines commensurable in square only.
ὥστε ἡ ΔΗ ἄλογός ἐστιν.
So that DH is irrational.
ῥητὴ δὲ ἡ ΔΕ· τὸ δὲ ὑπὸ ἀλόγου καὶ ῥητῆς περιεχόμενον ὀρθογώνιον ἄλογόν ἐστιν· ἄλογον ἄρα ἐστὶ τὸ ΔΖ χωρίον, καὶ ἡ δυναμένη ἄλογός ἐστιν.
And DE is rational; and the rectangle contained by an irrational and a rational straight line is irrational; therefore the area DZ is irrational, and the straight line which is equal in square to it is irrational.
δύναται δὲ τὸ ΔΖ ἡ ΑΓ· ἄλογος ἄρα ἐστὶν ἡ ΑΓ, καλείσθω δὲ ἐκ δύο μέσων δευτέρα. ὅπερ ἔδει δεῖξαι.
And AC is equal in square to DZ; therefore AC is irrational, and let it be called the second bimedial; which was to be proved.
§10.prop1.39ἐὰν δύο εὐθεῖαι δυνάμει ἀσύμμετροι συντεθῶσι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων ῥητόν, τὸ δʼ ὑπʼ αὐτῶν μέσον, ἡ ὅλη εὐθεῖα ἄλογός ἐστιν, καλείσθω δὲ μείζων.
If two straight lines incommensurable in square be added together making the sum of the squares on them rational, but the rectangle contained by them medial, the whole straight line is irrational, and let it be called major.
Συγκείσθωσαν γὰρ δύο εὐθεῖαι δυνάμει ἀσύμμετροι αἱ ΑΒ, ΒΓ ποιοῦσαι τὰ προκείμενα· λέγω, ὅτι ἄλογός ἐστιν ἡ ΑΓ.
ἐπεὶ γὰρ τὸ ὑπὸ τῶν ΑΒ, ΒΓ μέσον ἐστίν, καὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ μέσον ἐστίν.
For let two straight lines AB, BC incommensurable in square be added together making the proposed conditions; I say that AC is irrational. For since the rectangle contained by AB, BC is medial, twice the rectangle contained by AB, BC is also medial.
τὸ δὲ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ ῥητόν· ἀσύμμετρον ἄρα ἐστὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ· ὥστε καὶ τὰ ἀπὸ τῶν ΑΒ, ΒΓ μετὰ τοῦ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, ὅπερ ἐστὶ τὸ ἀπὸ τῆς ΑΓ, ἀσύμμετρόν ἐστι τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ·
But the sum of the squares on AB, BC is rational; therefore twice the rectangle contained by AB, BC is incommensurable with the sum of the squares on AB, BC; so that the sum of the squares on AB, BC with twice the rectangle contained by AB, BC, which is the square on AC, is also incommensurable with the sum of the squares on AB, BC.
ἄλογον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΓ. ὥστε καὶ ἡ ΑΓ ἄλογός ἐστιν, καλείσθω δὲ μείζων. ὅπερ ἔδει δεῖξαι.
Therefore the square on AC is irrational. So that AC is also irrational, and let it be called major; which was to be proved.