§10.prop1.32εὑρεῖν δύο μέσας δυνάμει μόνον συμμέτρους μέσον περιεχούσας, ὥστε τὴν μείζονα τῆς ἐλάσσονος μεῖζον δύνασθαι τῷ ἀπὸ συμμέτρου ἑαυτῇ.
To find two medial straight lines commensurable in square only containing a medial area, so that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater.
Ἐκκείσθωσαν τρεῖς ῥηταὶ δυνάμει μόνον σύμμετροι αἱ α, Β, Γ, ὥστε τὴν Α τῆς Γ μεῖζον δύνασθαι τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ τῷ μὲν ὑπὸ τῶν Α, Β ἴσον ἔστω τὸ ἀπὸ τῆς Δ. μέσον ἄρα τὸ ἀπὸ τῆς Δ· καὶ ἡ Δ ἄρα μέση ἐστίν.
Let three rational straight lines commensurable in square only A, B, C be set out, so that the square on A is greater than the square on C by the square on a straight line commensurable in length with A. And let the square on D be equal to the rectangle contained by A, B; therefore the square on D is medial, and D is also medial.
τῷ δὲ ὑπὸ τῶν Β, Γ ἴσον ἔστω τὸ ὑπὸ τῶν Δ, Ε. καὶ ἐπεί ἐστιν ὡς τὸ ὑπὸ τῶν Α, Β πρὸς τὸ ὑπὸ τῶν Β, Γ, οὕτως ἡ Α πρὸς τὴν Γ, ἀλλὰ τῷ μὲν ὑπὸ τῶν Α, Β ἴσον ἐστὶ τὸ ἀπὸ τῆς Δ, τῷ δὲ ὑπὸ τῶν Β, Γ ἴσον τὸ ὑπὸ τῶν Δ, Ε, ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Γ, οὕτως τὸ ἀπὸ τῆς Δ πρὸς τὸ ὑπὸ τῶν Δ, Ε. ὡς δὲ τὸ ἀπὸ τῆς Δ πρὸς τὸ ὑπὸ τῶν Δ, Ε, οὕτως ἡ Δ πρὸς τὴν Ε· καὶ ὡς ἄρα ἡ Α πρὸς τὴν Γ, οὕτως ἡ Δ πρὸς τὴν Ε·
And let the rectangle contained by D, E be equal to the rectangle contained by B, C. And since, as the rectangle contained by A, B is to the rectangle contained by B, C, so is A to C, but the square on D is equal to the rectangle contained by A, B, and the rectangle contained by D, E is equal to the rectangle contained by B, C, therefore, as A is to C, so is the square on D to the rectangle contained by D, E. But, as the square on D is to the rectangle contained by D, E, so is D to E; therefore also, as A is to C, so is D to E.
σύμμετρος δὲ ἡ Α τῇ Γ δυνάμει. σύμμετρος ἄρα καὶ ἡ Δ τῇ Ε δυνάμει μόνον.
But A is commensurable in square only with C; therefore D is also commensurable in square only with E.
μέση δὲ ἡ Δ· μέση ἄρα καὶ ἡ Ε. καὶ ἐπεί ἐστιν ὡς ἡ Α πρὸς τὴν Γ, ἡ Δ πρὸς τὴν Ε, ἡ δὲ Α τῆς Γ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ ἡ Δ ἄρα τῆς Ε μεῖζον δυνήσεται τῷ ἀπὸ συμμέτρου ἑαυτῇ.
And D is medial; therefore E is also medial. And since, as A is to C, so is D to E, and the square on A is greater than the square on C by the square on a straight line commensurable with A, therefore the square on D will also be greater than the square on E by the square on a straight line commensurable with D.
λέγω δή, ὅτι καὶ μέσον ἐστὶ τὸ ὑπὸ τῶν Δ, Ε. ἐπεὶ γὰρ ἴσον ἐστὶ τὸ ὑπὸ τῶν Β, Γ τῷ ὑπὸ τῶν Δ, Ε, μέσον δὲ τὸ ὑπὸ τῶν Β, Γ, μέσον ἄρα καὶ τὸ ὑπὸ τῶν Δ, Ε.
Εὕρηνται ἄρα δύο μέσαι δυνάμει μόνον σύμμετροι αἱ Δ, Ε μέσον περιέχουσαι, ὥστε τὴν μείζονα τῆς ἐλάσσονος μεῖζον δύνασθαι τῷ ἀπὸ συμμέτρου ἑαυτῇ.
I say then, that the rectangle contained by D, E is also medial. For since the rectangle contained by B, C is equal to the rectangle contained by D, E, and the rectangle contained by B, C is medial, therefore the rectangle contained by D, E is also medial. Therefore two medial straight lines D, E commensurable in square only containing a medial area have been found, so that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater.
ὁμοίως δὴ πάλιν δειχθήσεται καὶ τῷ ἀπὸ ἀσυμμέτρου, ὅταν ἡ Α τῆς Γ μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ.
And similarly it will also be shown for the square on an incommensurable straight line, when the square on A is greater than the square on C by the square on a straight line incommensurable with A.
λῆμμα
ἔστω τρίγωνον ὀρθογώνιον τὸ ΑΒΓ ὀρθὴν ἔχον τὴν α, καὶ ἤχθω κάθετος ἡ ΑΔ· λέγω, ὅτι τὸ μὲν ὑπὸ τῶν ΓΒΔ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΒΑ, τὸ δὲ ὑπὸ τῶν ΒΓΔ ἴσον τῷ ἀπὸ τῆς ΓΑ, καὶ τὸ ὑπὸ τῶν ΒΔ, ΔΓ ἴσον τῷ ἀπὸ τῆς ΑΔ, καὶ ἔτι τὸ ὑπὸ τῶν ΒΓ, ΑΔ ἴσον τῷ ὑπὸ τῶν ΒΑ, ΑΓ.
καὶ πρῶτον, ὅτι τὸ ὑπὸ τῶν ΓΒΔ ἴσον τῷ ἀπὸ τῆς ΒΑ.
ἐπεὶ γὰρ ἐν ὀρθογωνίῳ τριγώνῳ ἀπὸ τῆς ὀρθῆς γωνίας ἐπὶ τὴν βάσιν κάθετος ἦκται ἡ ΑΔ, τὰ ΑΒΔ, ΑΔΓ ἄρα τρίγωνα ὅμοιά ἐστι τῷ τε ὅλῳ τῷ ΑΒΓ καὶ ἀλλήλοις.
Lemma Let ABG be a right-angled triangle having the angle A right, and let AD be drawn perpendicular; I say that the rectangle contained by GB, BD is equal to the square on BA, the rectangle contained by BG, GD is equal to the square on GA, the rectangle contained by BD, DG is equal to the square on AD, and further the rectangle contained by BG, AD is equal to the rectangle contained by BA, AG. And first, that the rectangle contained by GB, BD is equal to the square on BA. For since in a right-angled triangle the perpendicular AD has been drawn from the right angle to the base, therefore the triangles ABD, ADG are similar both to the whole ABG and to one another.
καὶ ἐπεὶ ὅμοιόν ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΑΒΔ τριγώνῳ, ἔστιν ἄρα ὡς ἡ ΓΒ πρὸς τὴν ΒΑ, οὕτως ἡ ΒΑ πρὸς τὴν ΒΔ· τὸ ἄρα ὑπὸ τῶν ΓΒΔ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ.
διὰ τὰ αὐτὰ δὴ καὶ τὸ ὑπὸ τῶν ΒΓΔ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΓ.
καὶ ἐπεί, ἐὰν ἐν ὀρθογωνίῳ τριγώνῳ ἀπὸ τῆς ὀρθῆς γωνίας ἐπὶ τὴν βάσιν κάθετος ἀχθῇ, ἡ ἀχθεῖσα τῶν τῆς βάσεως τμημάτων μέση ἀνάλογόν ἐστιν, ἔστιν ἄρα ὡς ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΑΔ πρὸς τὴν ΔΓ· τὸ ἄρα ὑπὸ τῶν ΒΔ, ΔΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΑ.
λέγω, ὅτι καὶ τὸ ὑπὸ τῶν ΒΓ, ΑΔ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΒΑ, ΑΓ. ἐπεὶ γάρ, ὡς ἔφαμεν, ὅμοιόν ἐστι τὸ ΑΒΓ τῷ ΑΒΔ, ἔστιν ἄρα ὡς ἡ ΒΓ πρὸς τὴν ΓΑ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΔ. τὸ ἄρα ὑπὸ τῶν ΒΓ, ΑΔ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΒΑ, ΑΓ· ὅπερ ἔδει δεῖξαι.
And since the triangle ABG is similar to the triangle ABD, therefore, as GB is to BA, so is BA to BD; therefore the rectangle contained by GB, BD is equal to the square on AB. For the same reasons, the rectangle contained by BG, GD is also equal to the square on AC. And since, if in a right-angled triangle a perpendicular is drawn from the right angle to the base, the perpendicular so drawn is a mean proportional between the segments of the base, therefore, as BD is to DA, so is AD to DG; therefore the rectangle contained by BD, DG is equal to the square on DA. I say that the rectangle contained by BG, AD is also equal to the rectangle contained by BA, AC. For since, as we said, ABG is similar to ABD, therefore, as BG is to GA, so is BA to AD. Therefore the rectangle contained by BG, AD is equal to the rectangle contained by BA, AC; which was to be proved.