Classification of incommensurables: Book 10 Proposition 96
Translations
Ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς ἕκτης, ἡ τὸ χωρίον δυναμένη μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν. Χωρίον γὰρ τὸ ΑΒ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΓ καὶ ἀποτομῆς ἕκτης τῆς ΑΔ: λέγω, ὅτι ἡ τὸ ΑΒ χωρίον δυναμένη [ἡ] μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν. 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. 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. 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. 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. 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. ἔστω αὐτῶν διάμετρος ἡ ΟΡ, καὶ καταγεγράφθω τὸ σχῆμα. ὁμοίως δὴ τοῖς ἐπάνω δείξομεν, ὅτι ἡ ΛΝ δύναται τὸ ΑΒ χωρίον. Λέγω, ὅτι ἡ ΛΝ [ἡ] μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν. 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. πάλιν, ἐπεὶ μέσον ἐδείχθη τὸ ΔΚ καί ἐστιν ἴσον τῷ δὶς ὑπὸ τῶν ΛΟ, ΟΝ, καὶ τὸ δὶς ὑπὸ τῶν ΛΟ, ΟΝ μέσον ἐστίν. καὶ ἐπεὶ ἀσύμμετρον ἐδείχθη τὸ ΑΚ τῷ ΔΚ, ἀσύμμετρα [ἄρα] 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 ἀπ' αὐτῶν τετραγώνων μέσον καὶ τὸ δὶς ὑπ' αὐτῶν μέσον ἔτι τε τὰ ἀπ' αὐτῶν τετράγωνα ἀσύμμετρα τῷ δὶς ὑπ' αὐτῶν. ἡ ἄρα ΛΝ ἄλογός ἐστιν ἡ καλουμένη μετὰ μέσου μέσον τὸ ὅλον ποιοῦσα: καὶ δύναται τὸ ΑΒ χωρίον. Ἡ ἄρα τὸ χωρίον δυναμένη μετὰ μέσου μέσον τὸ ὅλον ποιοῦσά ἐστιν: ὅπερ ἔδει δεῖξαι.
If an area be contained by a rational straight line and a sixth apotome, the side of the area is a straight line which produces with a medial area a medial whole. For let the area AB be contained by the rational straight line AC and the sixth apotome AD; I say that the side of the area AB is a straight line which produces with a medial area a medial whole. For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, neither of them is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG. [X. Deff. III. 6] Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG, therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18] Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF, FG; therefore AF is incommensurable in length with FG. But, as AF is to FG, so is AI to FK. [VI. 1] therefore AI is incommensurable with FK. [X. 11] And, since AG, AC are rational straight lines commensurable in square only, AK is medial. [X. 21] Again, since AC, DG are rational straight lines and incommensurable in length, DK is also medial. [X. 21] Now, since AG, GD are commensurable in square only, therefore AG is incommensurable in length with GD. But, as AG is to GD, so is AK to KD; [VI. 1] therefore AK is incommensurable with KD. [X. 11] Now let the square LM be constructed equal to AI, and let NO equal to FK, and about the same angle, be subtracted; therefore the squares LM, NO are about the same diameter. [VI. 26] Let PR be their diameter, and let the figure be drawn. Then in manner similar to the above we can prove that LN is the side of the area AB. I say that LN is a straight line which produces with a medial area a medial whole. For, since AK was proved medial and is equal to the squares on LP, PN, therefore the sum of the squares on LP, PN is medial. Again, since DK was proved medial and is equal to twice the rectangle LP, PN, twice the rectangle LP, PN is also medial. And, since AK was proved incommensurable with DK, the squares on LP, PN are also incommensurable with twice the rectangle LP, PN. And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN; therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them medial, twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them. Therefore LN is the irrational straight line called that which produces with a medial area a medial whole; [X. 78] and it is the side of the area AB.