§10.prop1.28#1μέσας εὑρεῖν δυνάμει μόνον συμμέτρους μέσον περιεχούσας.
To find medial straight lines commensurable in square only which contain a medial rectangle.
Ἐκκείσθωσαν ῥηταὶ δυνάμει μόνον σύμμετροι αἱ Α, Β, Γ, καὶ εἰλήφθω τῶν Α, Β μέση ἀνάλογον ἡ Δ, καὶ γεγονέτω ὡς ἡ Β πρὸς τὴν Γ, ἡ Δ πρὸς τὴν Ε.
ἐπεὶ αἱ Α, Β ῥηταί εἰσι δυνάμει μόνον σύμμετροι, τὸ ἄρα ὑπὸ τῶν Α, Β, τουτέστι τὸ ἀπὸ τῆς Δ, μέσον ἐστίν.
Let there be set out rational straight lines A, B, C commensurable in square only, and let D be taken as a mean proportional between A, B, and let it be made: as B is to C, so is D to E. Since A, B are rational straight lines commensurable in square only, therefore the rectangle contained by A, B, that is, the square on D, is medial.
μέση ἄρα ἡ Δ. καὶ ἐπεὶ αἱ Β, Γ δυνάμει μόνον εἰσὶ σύμμετροι, καί ἐστιν ὡς ἡ Β πρὸς τὴν Γ, ἡ Δ πρὸς τὴν Ε, καὶ αἱ Δ, Ε ἄρα δυνάμει μόνον εἰσὶ σύμμετροι.
Therefore D is medial. And since B, C are commensurable in square only, and as B is to C, so is D to E, therefore D, E are also commensurable in square only.
μέση δὲ ἡ Δ· μέση ἄρα καὶ ἡ Ε·
But D is medial; therefore E is also medial.
αἱ Δ, Ε ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι.
Therefore D, E are medial straight lines commensurable in square only.
λέγω δή, ὅτι καὶ μέσον περιέχουσιν.
I say then that they also contain a medial rectangle.
ἐπεὶ γάρ ἐστιν ὡς ἡ Β πρὸς τὴν Γ, ἡ Δ πρὸς τὴν Ε, ἐναλλὰξ ἄρα ὡς ἡ Β πρὸς τὴν Δ, ἡ Γ πρὸς τὴν Ε. ὡς δὲ ἡ Β πρὸς τὴν Δ, ἡ Δ πρὸς τὴν Α·
For since as B is to C, so is D to E, therefore, alternately, as B is to D, so is C to E.
καὶ ὡς ἄρα ἡ Δ πρὸς τὴν Α, ἡ Γ πρὸς τὴν Ε· τὸ ἄρα ὑπὸ τῶν Α, Γ ἴσον ἐστὶ τῷ ὑπὸ τῶν Δ, Ε. μέσον δὲ τὸ ὑπὸ τῶν Α, Γ· μέσον ἄρα καὶ τὸ ὑπὸ τῶν Δ, Ε.
Εὕρηνται ἄρα μέσαι δυνάμει μόνον σύμμετροι μέσον περιέχουσαι· ὅπερ ἔδει δεῖξαι.
But as B is to D, so is D to A; therefore also as D is to A, so is C to E; therefore the rectangle contained by A, C is equal to the rectangle contained by D, E. But the rectangle contained by A, C is medial; therefore the rectangle contained by D, E is also medial. Therefore medial straight lines commensurable in square only which contain a medial rectangle have been found; which was to be proved.
λῆμμα
εὑρεῖν δύο τετραγώνους ἀριθμούς, ὥστε καὶ τὸν συγκείμενον ἐξ αὐτῶν εἶναι τετράγωνον.
Lemma To find two square numbers such that their sum is also a square.
Ἐκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΒ, ΒΓ, ἔστωσαν δὲ ἤτοι ἄρτιοι ἢ περιττοί.
Let there be set out two numbers AB, BC, and let them be either both even or both odd.
καὶ ἐπεί, ἐάν τε ἀπὸ ἀρτίου ἄρτιος ἀφαιρεθῇ, ἐάν τε ἀπὸ περισσοῦ περισσός, ὁ λοιπὸς ἄρτιός ἐστιν, ὁ λοιπὸς ἄρα ὁ ΑΓ ἄρτιός ἐστιν.
And since, whether an even number is subtracted from an even number, or an odd number from an odd number, the remainder is even, therefore the remainder AC is even.
τετμήσθω ὁ ΑΓ δίχα κατὰ τὸ Δ. ἔστωσαν δὲ καὶ οἱ ΑΒ, ΒΓ ἤτοι ὅμοιοι ἐπίπεδοι ἢ τετράγωνοι, οἳ καὶ αὐτοὶ ὅμοιοί εἰσιν ἐπίπεδοι· ὁ ἄρα ἐκ τῶν ΑΒ, ΒΓ μετὰ τοῦ ἀπὸ ΓΔ τετραγώνου ἴσος ἐστὶ τῷ ἀπὸ τοῦ ΒΔ τετραγώνῳ.
Let AC be bisected at D. And let AB, BC also be either similar plane numbers or square numbers, which themselves are also similar plane numbers; therefore the product of AB, BC together with the square on CD is equal to the square on BD.
καί ἐστι τετράγωνος ὁ ἐκ τῶν ΑΒ, ΒΓ, ἐπειδήπερ ἐδείχθη, ὅτι, ἐὰν δύο ὅμοιοι ἐπίπεδοι πολλαπλασιάσαντες ἀλλήλους ποιῶσί τινα, ὁ γενόμενος τετράγωνός ἐστιν.
And the product of AB, BC is a square, because it has been proved that, if two similar plane numbers by multiplying one another make some number, the product is a square.
εὕρηνται ἄρα δύο τετράγωνοι ἀριθμοὶ ὅ τε ἐκ τῶν ΑΒ, ΒΓ καὶ ὁ ἀπὸ τοῦ ΓΔ, οἳ συντεθέντες ποιοῦσι τὸν ἀπὸ τοῦ ΒΔ τετράγωνον.
Therefore two square numbers have been found, namely the product of AB, BC and the square on CD, which when added together make the square on BD.
καὶ φανερόν, ὅτι εὕρηνται πάλιν δύο τετράγωνοι ὅ τε ἀπὸ τοῦ ΒΔ καὶ ὁ ἀπὸ τοῦ ΓΔ, ὥστε τὴν ὑπεροχὴν αὐτῶν τὸν ὑπὸ ΑΒ, ΒΓ εἶναι τετράγωνον, ὅταν οἱ ΑΒ, ΒΓ ὅμοιοι ὦσιν ἐπίπεδοι.
And it is manifest that, again, two squares have been found, namely the square on BD and the square on CD, such that their difference, the product of AB, BC, is a square, when AB, BC are similar plane numbers.
ὅταν δὲ μὴ ὦσιν ὅμοιοι ἐπίπεδοι, εὕρηνται δύο τετράγωνοι ὅ τε ἀπὸ τοῦ ΒΔ καὶ ὁ ἀπὸ τοῦ ΔΓ, ὧν ἡ ὑπεροχὴ ὁ ὑπὸ τῶν ΑΒ, ΒΓ οὐκ ἔστι τετράγωνος· ὅπερ ἔδει δεῖξαι.
But when they are not similar plane numbers, two squares have been found, namely the square on BD and the square on DC, of which the difference, the product of AB, BC, is not a square; which was to be proved.