§10.prop3.96ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς ἕκτης, ἡ τὸ χωρίον δυναμένη μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν.
If an area be contained by a rational straight line and a sixth apotome, the straight line producing the area is that which produces with a medial area a medial whole.
χωρίον γὰρ τὸ ΑΒ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΓ καὶ ἀποτομῆς ἕκτης τῆς ΑΔ· λέγω, ὅτι ἡ τὸ ΑΒ χωρίον δυναμένη μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν.
For let the area AB be contained by the rational straight line AC and the sixth apotome AD; I say that the straight line producing the area AB is that which produces with a medial area a medial whole.
ἔστω γὰρ τῇ ΑΔ προσαρμόζουσα ἡ ΔΗ· αἱ ἄρα ΑΗ, ΗΔ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ οὐδετέρα αὐτῶν σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΑΓ μήκει, ἡ δὲ ὅλη ἡ ΑΗ τῆς προσαρμοζούσης τῆς ΔΗ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει.
For let DH be the annex to AD; therefore AH, HD are rational straight lines commensurable in square only, and neither of them 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.
ἐπεὶ οὖν ἡ ΑΗ τῆς ΗΔ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει, ἐὰν ἄρα τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΔΗ ἴσον παρὰ τὴν ΑΗ παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς ἀσύμμετρα αὐτὴν διελεῖ.
Since therefore AH is greater in square than DH by the square on a straight line incommensurable in length with itself, 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, 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.
ὡς δὲ ἡ ΑΖ πρὸς τὴν ΖΗ, οὕτως ἐστὶ τὸ ΑΙ πρὸς τὸ ΖΚ· ἀσύμμετρον ἄρα ἐστὶ τὸ ΑΙ τῷ ΖΚ. καὶ ἐπεὶ αἱ ΑΗ, ΑΓ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, μέσον ἐστὶ τὸ ΑΚ. πάλιν, ἐπεὶ αἱ ΑΓ, ΔΗ ῥηταί εἰσι καὶ ἀσύμμετροι μήκει, μέσον ἐστὶ καὶ τὸ ΔΚ. ἐπεὶ οὖν αἱ ΑΗ, ΗΔ δυνάμει μόνον σύμμετροί εἰσιν, ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΗ τῇ ΗΔ μήκει.
And as AZ is to ZH, so is AI to ZK; therefore AI is incommensurable with ZK. And, since AH, AC are rational straight lines commensurable in square only, AK is medial. Again, since AC, DH are rational straight lines and incommensurable in length, DK is also medial. Since therefore AH, HD are commensurable in square only, therefore AH is incommensurable in length with HD.
ὡς δὲ ἡ ΑΗ πρὸς τὴν ΗΔ, οὕτως ἐστὶ τὸ ΑΚ πρὸς τὸ ΚΔ· ἀσύμμετρον ἄρα ἐστὶ τὸ ΑΚ τῷ ΚΔ. συνεστάτω οὖν τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, τῷ δὲ ΖΚ ἴσον ἀφῃρήσθω περὶ τὴν αὐτὴν γωνίαν τὸ ΝΞ· περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὰ ΛΜ, ΝΞ τετράγωνα.
And as AH is to HD, so is AK to KD; therefore AK is incommensurable with KD. 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; 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 τοῖς ἐπάνω δείξομεν, ὅτι ἡ ΛΝ δύναται τὸ ΑΒ χωρίον.
Similarly then to the above we shall prove that LN produces the area AB.
λέγω, ὅτι ἡ ΛΝ μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν.
I say that LN is that which produces with a medial 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 μέσον ἐδείχθη τὸ ΔΚ καί ἐστιν ἴσον τῷ δὶς ὑπὸ τῶν ΛΟ, ΟΝ, καὶ τὸ δὶς ὑπὸ τῶν ΛΟ, ΟΝ μέσον ἐστίν.
Again, since DK was proved medial and is equal to twice the rectangle contained by LO, ON, twice the rectangle contained by LO, ON is also medial.
καὶ ἐπεὶ ἀσύμμετρον ἐδείχθη τὸ ΑΚ τῷ ΔΚ, ἀσύμμετρα ἐστὶ καὶ τὰ ἀπὸ τῶν ΛΟ, ΟΝ τετράγωνα τῷ δὶς ὑπὸ τῶν ΛΟ, ΟΝ. καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ ΑΙ τῷ ΖΚ, ἀσύμμετρον ἄρα καὶ τὸ ἀπὸ τῆς ΛΟ τῷ ἀπὸ τῆς ΟΝ· αἱ ΛΟ, ΟΝ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων μέσον καὶ τὸ δὶς ὑπʼ αὐτῶν μέσον ἔτι τε τὰ ἀπʼ αὐτῶν τετράγωνα ἀσύμμετρα τῷ δὶς ὑπʼ αὐτῶν.
And, since AK was proved incommensurable with DK, the squares on LO, ON are also incommensurable with twice the rectangle contained by LO, ON. 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, and twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them.
ἡ ἄρα ΛΝ ἄλογός ἐστιν ἡ καλουμένη μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα· καὶ δύναται τὸ ΑΒ χωρίον.
Therefore the remainder LN is the irrational straight line called that which produces with a medial area a medial whole; and it produces the area AB.
ἡ ἄρα τὸ χωρίον δυναμένη μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν· ὅπερ ἔδει δεῖξαι.
Therefore the straight line producing the area is that which produces with a medial area a medial whole; which was to be proved.