§10.prop1.26μέσον μέσου οὐχ ὑπερέχει ῥητῷ.
A medial area does not exceed a medial area by a rational area.
εἰ γὰρ δυνατόν, μέσον τὸ ΑΒ μέσου τοῦ ΑΓ ὑπερεχέτω ῥητῷ τῷ ΔΒ, καὶ ἐκκείσθω ῥητὴ ἡ ΕΖ, καὶ τῷ ΑΒ ἴσον παρὰ τὴν ΕΖ παραβεβλήσθω παραλληλόγραμμον ὀρθογώνιον τὸ ΖΘ πλάτος ποιοῦν τὴν ΕΘ, τῷ δὲ ΑΓ ἴσον ἀφῃρήσθω τὸ ΖΗ· λοιπὸν ἄρα τὸ ΒΔ λοιπῷ τῷ ΚΘ ἐστιν ἴσον.
For, if possible, let the medial area AB exceed the medial area AC by the rational area DB, and let the rational straight line EZ be set out, and let there be applied to EZ the rectangular parallelogram ZΘ equal to AB, producing EΘ as breadth, and let ZH equal to AC be subtracted [from ZΘ]; therefore the remainder BD is equal to the remainder KΘ.
ῥητὸν δέ ἐστι τὸ ΔΒ· ῥητὸν ἄρα ἐστὶ καὶ τὸ ΚΘ. ἐπεὶ οὖν μέσον ἐστὶν ἑκάτερον τῶν ΑΒ, ΑΓ, καί ἐστι τὸ μὲν ΑΒ τῷ ΖΘ ἴσον, τὸ δὲ ΑΓ τῷ ΖΗ, μέσον ἄρα καὶ ἑκάτερον τῶν ΖΘ, ΖΗ. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται· ῥητὴ ἄρα ἐστὶν ἑκατέρα τῶν ΘΕ, ΕΗ καὶ ἀσύμμετρος τῇ ΕΖ μήκει.
But DB is rational; therefore KΘ is also rational. Since therefore each of the areas AB, AC is medial, and AB is equal to ZΘ, and AC to ZH, therefore each of ZΘ, ZH is also medial. And they are applied to the rational straight line EZ; therefore each of ΘE, EH is rational and incommensurable in length with EZ.
καὶ ἐπεὶ ῥητόν ἐστι τὸ ΔΒ καί ἐστιν ἴσον τῷ ΚΘ, ῥητὸν ἄρα ἐστὶ καὶ τὸ ΚΘ. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται· ῥητὴ ἄρα ἐστὶν ἡ ΗΘ καὶ σύμμετρος τῇ ΕΖ μήκει.
And since DB is rational and is equal to KΘ, therefore KΘ is also rational. And it is applied to the rational straight line EZ; therefore HΘ is rational and commensurable in length with EZ.
ἀλλὰ καὶ ἡ ΕΗ ῥητή ἐστι καὶ ἀσύμμετρος τῇ ΕΖ μήκει· ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΗ τῇ ΗΘ μήκει.
But EH is also rational and incommensurable in length with EZ; therefore EH is incommensurable in length with HΘ.
καί ἐστιν ὡς ἡ ΕΗ πρὸς τὴν ΗΘ, οὕτως τὸ ἀπὸ τῆς ΕΗ πρὸς τὸ ὑπὸ τῶν ΕΗ, ΗΘ· ἀσύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΕΗ τῷ ὑπὸ τῶν ΕΗ, ΗΘ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΕΗ σύμμετρά ἐστι τὰ ἀπὸ τῶν ΕΗ, ΗΘ τετράγωνα· ῥητὰ γὰρ ἀμφότερα· τῷ δὲ ὑπὸ τῶν ΕΗ, ΗΘ σύμμετρόν ἐστι τὸ δὶς ὑπὸ τῶν ΕΗ, ΗΘ· διπλάσιον γάρ ἐστιν αὐτοῦ· ἀσύμμετρα ἄρα ἐστὶ τὰ ἀπὸ τῶν ΕΗ, ΗΘ τῷ δὶς ὑπὸ τῶν ΕΗ, ΗΘ· καὶ συναμφότερα ἄρα τά τε ἀπὸ τῶν ΕΗ, ΗΘ καὶ τὸ δὶς ὑπὸ τῶν ΕΗ, ΗΘ, ὅπερ ἐστὶ τὸ ἀπὸ τῆς ΕΘ, ἀσύμμετρόν ἐστι τοῖς ἀπὸ τῶν ΕΗ, ΗΘ. ῥητὰ δὲ τὰ ἀπὸ τῶν ΕΗ, ΗΘ·
And as EH is to HΘ, so is the square on EH to the rectangle contained by EH, HΘ; therefore the square on EH is incommensurable with the rectangle contained by EH, HΘ. But the squares on EH, HΘ [together] are commensurable with the square on EH; for both are rational; and twice the rectangle contained by EH, HΘ is commensurable with the rectangle contained by EH, HΘ; for it is double of it; therefore the squares on EH, HΘ [together] are incommensurable with twice the rectangle contained by EH, HΘ; and therefore both together, namely the squares on EH, HΘ [together] and twice the rectangle contained by EH, HΘ, which is the square on EΘ, is incommensurable with the squares on EH, HΘ [together]. And the squares on EH, HΘ [together] are rational; therefore the square on EΘ is irrational.
ἄλογον ἄρα τὸ ἀπὸ τῆς ΕΘ. ἄλογος ἄρα ἐστὶν ἡ ΕΘ. ἀλλὰ καὶ ῥητή· ὅπερ ἐστὶν ἀδύνατον.
Therefore EΘ is irrational. But it is also rational; which is impossible.
μέσον ἄρα μέσου οὐχ ὑπερέχει ῥητῷ· ὅπερ ἔδει δεῖξαι.
Therefore a medial area does not exceed a medial area by a rational area; which was to be proved.
§10.prop1.27μέσας εὑρεῖν δυνάμει μόνον συμμέτρους ῥητὸν περιεχούσας.
To find medial straight lines commensurable in square only which contain a rational rectangle.
Ἐκκείσθωσαν δύο ῥηταὶ δυνάμει μόνον σύμμετροι αἱ Α, Β, καὶ εἰλήφθω τῶν Α, Β μέση ἀνάλογον ἡ Γ, καὶ γεγονέτω ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ.
καὶ ἐπεὶ αἱ Α, Β ῥηταί εἰσι δυνάμει μόνον σύμμετροι, τὸ ἄρα ὑπὸ τῶν Α, Β, τουτέστι τὸ ἀπὸ τῆς Γ, μέσον ἐστίν.
Let there be set out two rational straight lines A, B commensurable in square only, and let C be taken as a mean proportional between A, B, and let it be made: as A is to B, so is C to D. And since A, B are rational straight lines commensurable in square only, therefore the rectangle contained by A, B, that is, the square on C, is medial.
μέση ἄρα ἡ Γ. καὶ ἐπεί ἐστιν ὡς ἡ Α πρὸς τὴν Β, ἡ Γ πρὸς τὴν Δ, αἱ δὲ Α, Β δυνάμει μόνον σύμμετροι, καὶ αἱ Γ, Δ ἄρα δυνάμει μόνον εἰσὶ σύμμετροι.
Therefore C is medial. And since as A is to B, so is C to D, and A, B are commensurable in square only, therefore C, D are also commensurable in square only.
καί ἐστι μέση ἡ Γ· μέση ἄρα καὶ ἡ Δ. αἱ Γ, Δ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι.
And C is medial; therefore D is also medial. Therefore C, D are medial straight lines commensurable in square only.
λέγω, ὅτι καὶ ῥητὸν περιέχουσιν.
I say that they also contain a rational rectangle.
ἐπεὶ γάρ ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ, ἐναλλὰξ ἄρα ἐστὶν ὡς ἡ Α πρὸς τὴν Γ, ἡ Β πρὸς τὴν Δ. ἀλλʼ ὡς ἡ Α πρὸς τὴν Γ, ἡ Γ πρὸς τὴν Β· καὶ ὡς ἄρα ἡ Γ πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Δ·
For since as A is to B, so is C to D, therefore, alternately, as A is to C, so is B to D. But as A is to C, so is C to B; therefore also as C is to B, so is B to D; therefore the rectangle contained by C, D is equal to the square on B.
τὸ ἄρα ὑπὸ τῶν Γ, Δ ἴσον ἐστὶ τῷ ἀπὸ τῆς Β. ῥητὸν δὲ τὸ ἀπὸ τῆς Β· ῥητὸν ἄρα καὶ τὸ ὑπὸ τῶν Γ, Δ.
Εὕρηνται ἄρα μέσαι δυνάμει μόνον σύμμετροι ῥητὸν περιέχουσαι· ὅπερ ἔδει δεῖξαι.
But the square on B is rational; therefore the rectangle contained by C, D is also rational. Therefore medial straight lines commensurable in square only which contain a rational rectangle have been found; which was to be proved.