§10.prop1.45ἡ μείζων κατὰ τὸ αὐτὸ μόνον σημεῖον διαιρεῖται.
The major straight line is divided at the same point only.
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 δʼ ὑπὸ τῶν ΑΓ, ΓΒ μέσον·
For let the major straight line AB be divided at C, so that AC, CB are incommensurable in square, making the sum of the squares on AC, CB rational, but the rectangle contained by AC, CB medial.
λέγω, ὅτι ἡ ΑΒ κατʼ ἄλλο σημεῖον οὐ διαιρεῖται.
I say that AB is not divided at another point.
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 δʼ ὑπʼ αὐτῶν μέσον.
For, if possible, let it also be divided at D, so that AD, DB are also incommensurable in square, making the sum of the squares on AD, DB rational, but the rectangle contained by them medial.
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.
And since, by that which the sum of the squares on AC, CB differs from the sum of the squares on AD, DB, by this also twice the rectangle contained by AD, DB differs from twice the rectangle contained by AC, CB, but the sum of the squares on AC, CB exceeds the sum of the squares on AD, DB by a rational area; for both are rational; therefore twice the rectangle contained by AD, DB also exceeds twice the rectangle contained by AC, CB by a rational area, though they are medial; which is impossible.
οὐκ ἄρα ἡ μείζων κατʼ ἄλλο καὶ ἄλλο σημεῖον διαιρεῖται· κατὰ τὸ αὐτὸ ἄρα μόνον διαιρεῖται· ὅπερ ἔδει δεῖξαι.
Therefore the major straight line is not divided at different points; therefore it is divided at the same point only; which was to be proved.
§10.prop1.46ἡ ῥητὸν καὶ μέσον δυναμένη καθʼ ἓν μόνον σημεῖον διαιρεῖται.
The side of a rational plus a medial area is divided at one point only.
ἔστω ῥητὸν καὶ μέσον δυναμένη ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, ὥστε τὰς ΑΓ, ΓΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ μέσον, τὸ δὲ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ῥητόν·
For let the side of a rational plus a medial area AB be divided at C, so that AC, CB are incommensurable in square, making the sum of the squares on AC, CB medial, but twice the rectangle contained by AC, CB rational.
λέγω, ὅτι ἡ ΑΒ κατʼ ἄλλο σημεῖον οὐ διαιρεῖται.
I say that AB is not divided at another point.
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 μέσον, τὸ δὲ δὶς ὑπὸ τῶν ΑΔ, ΔΒ ῥητόν.
For, if possible, let it also be divided at D, so that AD, DB are also incommensurable in square, making the sum of the squares on AD, DB medial, but twice the rectangle contained by AD, DB rational.
ἐπεὶ οὖν, ᾧ διαφέρει τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ τοῦ δὶς ὑπὸ τῶν ΑΔ, ΔΒ, τούτῳ διαφέρει καὶ τὰ ἀπὸ τῶν ΑΔ, ΔΒ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, τὸ δὲ δὶς ὑπὸ τῶν ΑΓ, ΓΒ τοῦ δὶς ὑπὸ τῶν ΑΔ, ΔΒ ὑπερέχει ῥητῷ, καὶ τὰ ἀπὸ τῶν ΑΔ, ΔΒ ἄρα τῶν ἀπὸ τῶν ΑΓ, ΓΒ ὑπερέχει ῥητῷ μέσα ὄντα· ὅπερ ἐστὶν ἀδύνατον.
Since then, by that which twice the rectangle contained by AC, CB differs from twice the rectangle contained by AD, DB, by this also the sum of the squares on AD, DB differs from the sum of the squares on AC, CB, but twice the rectangle contained by AC, CB exceeds twice the rectangle contained by AD, DB by a rational area; therefore the sum of the squares on AD, DB also exceeds the sum of the squares on AC, CB by a rational area, though they are medial; which is impossible.
οὐκ ἄρα ἡ ῥητὸν καὶ μέσον δυναμένη κατʼ ἄλλο καὶ ἄλλο σημεῖον διαιρεῖται.
Therefore the side of a rational plus a medial area is not divided at different points.
κατὰ ἓν ἄρα σημεῖον διαιρεῖται· ὅπερ ἔδει δεῖξαι.
Therefore it is divided at one point only; which was to be proved.
§10.prop1.47ἡ δύο μέσα δυναμένη καθʼ ἓν μόνον σημεῖον διαιρεῖται.
The side of the sum of two medial areas is divided at one point only.
ἔστω ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, ὥστε τὰς ΑΓ, ΓΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας τό τε συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ μέσον καὶ τὸ ὑπὸ τῶν ΑΓ, ΓΒ μέσον καὶ ἔτι ἀσύμμετρον τῷ συγκειμένῳ ἐκ τῶν ἀπʼ αὐτῶν.
For let AB be divided at C, so that AC, CB are incommensurable in square, making the sum of the squares on AC, CB medial and the rectangle contained by AC, CB medial, and further incommensurable with the sum of the squares on them.
λέγω, ὅτι ἡ ΑΒ κατʼ ἄλλο σημεῖον οὐ διαιρεῖται ποιοῦσα τὰ προκείμενα.
I say that AB is not divided at another point making the aforesaid.
εἰ γὰρ δυνατόν, διῃρήσθω κατὰ τὸ Δ, ὥστε πάλιν δηλονότι τὴν ΑΓ τῇ ΔΒ μὴ εἶναι τὴν αὐτήν, ἀλλὰ μείζονα καθʼ ὑπόθεσιν τὴν ΑΓ, καὶ ἐκκείσθω ῥητὴ ἡ ΕΖ, καὶ παραβεβλήσθω παρὰ τὴν ΕΖ τοῖς μὲν ἀπὸ τῶν ΑΓ, ΓΒ ἴσον τὸ ΕΗ, τῷ δὲ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἴσον τὸ ΘΚ· ὅλον ἄρα τὸ ΕΚ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ.
For, if possible, let it be divided at D, so that again manifestly AC is not the same with DB, but let AC be greater by hypothesis, and let the rational straight line EF be set out, and let there be applied to EF, EH equal to the sum of the squares on AC, CB, and TH equal to twice the rectangle contained by AC, CB; therefore the whole EK is equal to the square on AB.
πάλιν δὴ παραβεβλήσθω παρὰ τὴν ΕΖ τοῖς ἀπὸ τῶν ΑΔ, ΔΒ ἴσον τὸ ΕΛ· λοιπὸν ἄρα τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ λοιπῷ τῷ ΜΚ ἴσον ἐστίν.
Again, let there be applied to EF, EL equal to the sum of the squares on AD, DB; therefore the remainder twice the rectangle contained by AD, DB is equal to the remainder MK.
καὶ ἐπεὶ μέσον ὑπόκειται τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, μέσον ἄρα ἐστὶ καὶ τὸ ΕΗ. καὶ παρὰ ῥητὴν τὴν ΕΖ παράκειται·
And, since the sum of the squares on AC, CB is assumed to be medial, EH is also medial.
ῥητὴ ἄρα ἐστὶν ἡ ΘΕ καὶ ἀσύμμετρος τῇ ΕΖ μήκει.
And it is applied to the rational straight line EF; therefore TE is rational and incommensurable in length with EF.
διὰ τὰ αὐτὰ δὴ καὶ ἡ ΘΝ ῥητή ἐστι καὶ ἀσύμμετρος τῇ ΕΖ μήκει.
For the same reason then, TN is also rational and incommensurable in length with EF.
καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, καὶ τὸ ΕΗ ἄρα τῷ ΗΝ ἀσύμμετρόν ἐστιν· ὥστε καὶ ἡ ΕΘ τῇ ΘΝ ἀσύμμετρός ἐστιν.
And, since the sum of the squares on AC, CB is incommensurable with twice the rectangle contained by AC, CB, EH is also incommensurable with HN; so that ET is also incommensurable with TN.
καί εἰσι ῥηταί· αἱ ΕΘ, ΘΝ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι·
And they are rational; therefore ET, TN are rational straight lines commensurable in square only; therefore EN is a binomial straight line divided at T.
ἡ ΕΝ ἄρα ἐκ δύο ὀνομάτων ἐστὶ διῃρημένη κατὰ τὸ Θ. ὁμοίως δὴ δείξομεν, ὅτι καὶ κατὰ τὸ Μ διῄρηται.
In the same way then, we shall prove that it is also divided at M.
καὶ οὐκ ἔστιν ἡ ΕΘ τῇ ΜΝ ἡ αὐτή· ἡ ἄρα ἐκ δύο ὀνομάτων κατʼ ἄλλο καὶ ἄλλο σημεῖον διῄρηται· ὅπερ ἐστὶν ἄτοπον.
And ET is not the same with MN; therefore the binomial straight line is divided at different points; which is absurd.
οὐκ ἄρα ἡ δύο μέσα δυναμένη κατʼ ἄλλο καὶ ἄλλο σημεῖον διαιρεῖται· καθʼ ἓν ἄρα μόνον διαιρεῖται.
Therefore the side of the sum of two medial areas is not divided at different points; therefore it is divided at one point only.