Ἡ μείζων κατὰ τὸ αὐτὸ μόνον σημεῖον διαιρεῖται. Ἔστω μείζων ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, ὥστε τὰς ΑΓ, ΓΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ τετραγώνων ῥητόν, τὸ δ' ὑπὸ τῶν ΑΓ, ΓΒ μέσον: λέγω, ὅτι ἡ ΑΒ κατ' ἄλλο σημεῖον οὐ διαιρεῖται. Εἰ γὰρ δυνατόν, διῃρήσθω καὶ κατὰ τὸ Δ, ὥστε καὶ τὰς ΑΔ, ΔΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΔ, ΔΒ ῥητόν, τὸ δ' ὑπ' αὐτῶν μέσον. καὶ ἐπεί, ᾧ διαφέρει τὰ ἀπὸ τῶν ΑΓ, ΓΒ τῶν ἀπὸ τῶν ΑΔ, ΔΒ, τούτῳ διαφέρει καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, ἀλλὰ τὰ ἀπὸ τῶν ΑΓ, ΓΒ τῶν ἀπὸ τῶν ΑΔ, ΔΒ ὑπερέχει ῥητῷ: ῥητὰ γὰρ ἀμφότερα: καὶ τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ ἄρα τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ὑπερέχει ῥητῷ μέσα ὄντα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ μείζων κατ' ἄλλο καὶ ἄλλο σημεῖον διαιρεῖται: κατὰ τὸ αὐτὸ ἄρα μόνον διαιρεῖται: ὅπερ ἔδει δεῖξαι.

A major straight line is divided at one and the same point only. Let AB be a major straight line divided at C, so that AC, CB are incommensurable in square and make the sum of the squares on AC, CB rational, but the rectangle AC, CB medial; [X. 39] I say that AB is not so divided at another point. For, if possible, let it be divided at D also, so that AD, DB are also incommensurable in square and make the sum of the squares on AD, DB rational, but the rectangle contained by them medial. Then, since that by which the squares on AC, CB differ from the squares on AD, DB is also that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB, while the squares on AC, CB exceed the squares on AD, DB by a rational areafor both are rational therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area, though they are medial: which is impossible. [X. 26]