§10.prop3.95ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς πέμπτης, ἡ τὸ χωρίον δυναμένη μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά ἐστιν.
If an area be contained by a rational straight line and a fifth apotome, the straight line producing the area is that which produces with a rational area a medial whole.
χωρίον γὰρ τὸ ΑΒ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΓ καὶ ἀποτομῆς πέμπτης τῆς ΑΔ· λέγω, ὅτι ἡ τὸ ΑΒ χωρίον δυναμένη μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά ἐστιν.
For let the area AB be contained by the rational straight line AC and the fifth apotome AD; I say that the straight line producing the area AB is that which produces with a rational area a medial whole.
ἔστω γὰρ τῇ ΑΔ προσαρμόζουσα ἡ ΔΗ· αἱ ἄρα ΑΗ, ΗΔ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ προςαρμόζουσα ἡ ΗΔ σύμμετρός ἐστι μήκει τῇ ἐκκειμένῃ ῥητῇ τῇ ΑΓ, ἡ δὲ ὅλη ἡ ΑΗ τῆς προσαρμοζούσης τῆς ΔΗ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ.
For let DH be the annex to AD; therefore AH, HD are rational straight lines commensurable in square only, and the annex DH is commensurable in length with the set-out rational straight line AC, but the whole AH is greater in square than the annex DH by the square on a straight line incommensurable in length with itself.
ἐὰν ἄρα τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΔΗ ἴσον παρὰ τὴν ΑΗ παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς ἀσύμμετρα αὐτὴν διελεῖ.
If therefore a rectangle equal to the fourth part of the square on DH be applied to AH, deficient by a square figure, it divides it into parts incommensurable in length.
τετμήσθω οὖν ἡ ΔΗ δίχα κατὰ τὸ Ε σημεῖον, καὶ τῷ ἀπὸ τῆς ΕΗ ἴσον παρὰ τὴν ΑΗ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ καὶ ἔστω τὸ ὑπὸ τῶν ΑΖ, ΖΗ· ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΖ τῇ ΖΗ μήκει.
Let then DH be bisected at the point E; and let there be applied to AH a rectangle equal to the square on EH, deficient by a square figure, and let it be the rectangle contained by AZ, ZH; therefore AZ is incommensurable in length with ZH.
καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΗ τῇ ΓΑ μήκει, καί εἰσιν ἀμφότεραι ῥηταί, μέσον ἄρα ἐστὶ τὸ ΑΚ. πάλιν, ἐπεὶ ῥητή ἐστιν ἡ ΔΗ καὶ σύμμετρος τῇ ΑΓ μήκει, ῥητόν ἐστι τὸ ΔΚ. συνεστάτω οὖν τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, τῷ δὲ ΖΚ ἴσον τετράγωνον ἀφῃρήσθω τὸ ΝΞ περὶ τὴν αὐτὴν γωνίαν τὴν ὑπὸ ΛΟΜ· περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὰ ΛΜ, ΝΞ τετράγωνα.
And, since AH is incommensurable in length with GA, and both are rational, therefore AK is medial. Again, since DH is rational and commensurable in length with AC, DK is rational. Let then the square LM be constructed equal to AI, and let the square NX, equal to ZK, be subtracted, having a common angle with the square LM, namely the angle contained by LOM; therefore the squares LM, NX are about the same diagonal.
ἔστω αὐτῶν διάμετρος ἡ ΟΡ, καὶ καταγεγράφθω τὸ σχῆμα.
Let OP be their diagonal, and let the figure be described.
ὁμοίως δὴ δείξομεν, ὅτι ἡ ΛΝ δύναται τὸ ΑΒ χωρίον.
Similarly then we shall prove that LN produces the area AB.
λέγω, ὅτι ἡ ΛΝ ἡ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά ἐστιν.
I say that LN is that which produces with a rational area a medial whole.
ἐπεὶ γὰρ μέσον ἐδείχθη τὸ ΑΚ καί ἐστιν ἴσον τοῖς ἀπὸ τῶν ΛΟ, ΟΝ, τὸ ἄρα συγκείμενον ἐκ τῶν ἀπὸ τῶν ΛΟ, ΟΝ μέσον ἐστίν.
For since AK was proved medial and is equal to the squares on LO, ON, therefore the sum of the squares on LO, ON is medial.
πάλιν, ἐπεὶ ῥητόν ἐστι τὸ ΔΚ καί ἐστιν ἴσον τῷ δὶς ὑπὸ τῶν ΛΟ, ΟΝ, καὶ αὐτὸ ῥητόν ἐστιν.
Again, since DK is rational and is equal to twice the rectangle contained by LO, ON, therefore twice the rectangle contained by LO, ON is also rational.
καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ ΑΙ τῷ ΖΚ, ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς ΛΟ τῷ ἀπὸ τῆς ΟΝ· αἱ ΛΟ, ΟΝ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων μέσον, τὸ δὲ δὶς ὑπʼ αὐτῶν ῥητόν.
And, since AI is incommensurable with ZK, therefore the square on LO is also incommensurable with the square on ON; therefore LO, ON are incommensurable in square, making the sum of the squares on them medial, but twice the rectangle contained by them rational.
ἡ λοιπὴ ἄρα ἡ ΛΝ ἄλογός ἐστιν ἡ καλουμένη μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσα· καὶ δύναται τὸ ΑΒ χωρίον.
Therefore the remainder LN is the irrational straight line called that which produces with a rational area a medial whole; and it produces the area AB.
ἡ τὸ ΑΒ ἄρα χωρίον δυναμένη μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσά ἐστιν· ὅπερ ἔδει δεῖξαι.
Therefore the straight line producing the area AB is that which produces with a rational area a medial whole; which was to be proved.