§10.prop3.93ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς τρίτης, ἡ τὸ χωρίον δυναμένη μέσης ἀποτομή ἐστι δευτέρα.
If an area be contained by a rational straight line and a third apotome, the straight line producing the area is a second apotome of a medial straight line.
χωρίον γὰρ τὸ ΑΒ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΓ καὶ ἀποτομῆς τρίτης τῆς ΑΔ· λέγω, ὅτι ἡ τὸ ΑΒ χωρίον δυναμένη μέσης ἀποτομή ἐστι δευτέρα.
For let the area AB be contained by the rational straight line AC and the third apotome AD; I say that the straight line producing the area AB is a second apotome of a medial straight line.
ἔστω γὰρ τῇ ΑΔ προσαρμόζουσα ἡ ΔΗ· αἱ ΑΗ, ΗΔ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ οὐδετέρα τῶν ΑΗ, ΗΔ σύμμετρός ἐστι μήκει τῇ ἐκκειμένῃ ῥητῇ τῇ ΑΓ, ἡ δὲ ὅλη ἡ ΑΗ τῆς προσαρμοζούσης τῆς ΔΗ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ.
For let DH be the annex to AD; therefore AH, HD are rational straight lines commensurable in square only, and neither of AH, HD 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 commensurable in length with itself.
ἐπεὶ οὖν ἡ ΑΗ τῆς ΗΔ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, ἐὰν ἄρα τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΔΗ ἴσον παρὰ τὴν ΑΗ παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς σύμμετρα αὐτὴν διελεῖ.
Since, then, AH is greater in square than DH by the square on a straight line commensurable 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 commensurable 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. καὶ ἤχθωσαν διὰ τῶν Ε, Ζ, Η σημείων τῇ ΑΓ παράλληλοι αἱ ΕΘ, ΖΙ, ΗΚ· σύμμετροι ἄρα εἰσὶν αἱ ΑΖ, ΖΗ·
Let then DH be bisected at 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. And let there be drawn through the points E, Z, H parallel to AC the straight lines EG, ZI, HK; therefore AZ, ZH are commensurable; therefore the area AI is also commensurable with ZK.
σύμμετρον ἄρα καὶ τὸ ΑΙ τῷ ΖΚ. καὶ ἐπεὶ αἱ ΑΖ, ΖΗ σύμμετροί εἰσι μήκει, καὶ ἡ ΑΗ ἄρα ἑκατέρᾳ τῶν ΑΖ, ΖΗ σύμμετρός ἐστι μήκει.
And since AZ, ZH are commensurable in length, therefore AH is also commensurable in length with each of AZ, ZH.
ῥητὴ δὲ ἡ ΑΗ καὶ ἀσύμμετρος τῇ ΑΓ μήκει· ὥστε καὶ αἱ ΑΖ, ΖΗ. ἑκάτερον ἄρα τῶν ΑΙ, ΖΚ μέσον ἐστίν.
And AH is rational and incommensurable in length with AC; so that AZ, ZH are also so. Therefore each of the areas AI, ZK is medial.
πάλιν, ἐπεὶ σύμμετρός ἐστιν ἡ ΔΕ τῇ ΕΗ μήκει, καὶ ἡ ΔΗ ἄρα ἑκατέρᾳ τῶν ΔΕ, ΕΗ σύμμετρός ἐστι μήκει.
Again, since DE is commensurable in length with EH, therefore DH is also commensurable in length with each of DE, EH.
ῥητὴ δὲ ἡ ΗΔ καὶ ἀσύμμετρος τῇ ΑΓ μήκει· ῥητὴ ἄρα καὶ ἑκατέρα τῶν ΔΕ, ΕΗ καὶ ἀσύμμετρος τῇ ΑΓ μήκει· ἑκάτερον ἄρα τῶν ΔΘ, ΕΚ μέσον ἐστίν.
And HD is rational and incommensurable in length with AC; therefore each of DE, EH is also rational and incommensurable in length with AC; therefore each of the areas DG, EK is medial.
καὶ ἐπεὶ αἱ ΑΗ, ΗΔ δυνάμει μόνον σύμμετροί εἰσιν, ἀσύμμετρος ἄρα ἐστὶ μήκει ἡ ΑΗ τῇ ΗΔ. ἀλλʼ ἡ μὲν ΑΗ τῇ ΑΖ σύμμετρός ἐστι μήκει, ἡ δὲ ΔΗ τῇ ΕΗ· ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΖ τῇ ΕΗ μήκει.
And since AH, HD are commensurable in square only, therefore AH is incommensurable in length with HD. But AH is commensurable in length with AZ, and DH with EH; therefore AZ is incommensurable in length with EH.
ὡς δὲ ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἐστὶ τὸ ΑΙ πρὸς τὸ ΕΚ· ἀσύμμετρον ἄρα ἐστὶ τὸ ΑΙ τῷ ΕΚ.
συνεστάτω οὖν τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, τῷ δὲ ΖΚ ἴσον ἀφῃρήσθω τὸ ΝΞ περὶ τὴν αὐτὴν γωνίαν ὂν τῷ ΛΜ· περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὰ ΛΜ, ΝΞ. ἔστω αὐτῶν διάμετρος ἡ ΟΡ, καὶ καταγεγράφθω τὸ σχῆμα.
And as AZ is to EH, so is AI to EK; therefore AI is incommensurable with EK. 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. Let OP be their diagonal, and let the figure be described.
ἐπεὶ οὖν τὸ ὑπὸ τῶν ΑΖ, ΖΗ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΗ, ἔστιν ἄρα ὡς ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἡ ΕΗ πρὸς τὴν ΖΗ. ἀλλʼ ὡς μὲν ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἐστὶ τὸ ΑΙ πρὸς τὸ ΕΚ· ὡς δὲ ἡ ΕΗ πρὸς τὴν ΖΗ, οὕτως ἐστὶ τὸ ΕΚ πρὸς τὸ ΖΚ· καὶ ὡς ἄρα τὸ ΑΙ πρὸς τὸ ΕΚ, οὕτως τὸ ΕΚ πρὸς τὸ ΖΚ·
Since, then, the rectangle contained by AZ, ZH is equal to the square on EH, therefore, as AZ is to EH, so is EH to ZH. But as AZ is to EH, so is AI to EK; and as EH is to ZH, so is EK to ZK; therefore also, as AI is to EK, so is EK to ZK; therefore EK is a mean proportional between AI, ZK.
τῶν ἄρα ΑΙ, ΖΚ μέσον ἀνάλογόν ἐστι τὸ ΕΚ. ἔστι δὲ καὶ τῶν ΛΜ, ΝΞ τετραγώνων μέσον ἀνάλογον τὸ ΜΝ· καί ἐστιν ἴσον τὸ μὲν ΑΙ τῷ ΛΜ, τὸ δὲ ΖΚ τῷ ΝΞ·
But MN is also a mean proportional between the squares LM, NX, and AI is equal to LM, and ZK to NX; therefore EK is also equal to MN.
καὶ τὸ ΕΚ ἄρα ἴσον ἐστὶ τῷ ΜΝ. ἀλλὰ τὸ μὲν ΜΝ ἴσον ἐστὶ τῷ ΛΞ, τὸ δὲ ΕΚ ἴσον τῷ ΔΘ·
But MN is equal to LX, and EK to DG; therefore the whole DK is equal to the gnomon UFX and NX.
καὶ ὅλον ἄρα τὸ ΔΚ ἴσον ἐστὶ τῷ ΥΦΧ γνώμονι καὶ τῷ ΝΞ. ἔστι δὲ καὶ τὸ ΑΚ ἴσον τοῖς ΛΜ, ΝΞ· λοιπὸν ἄρα τὸ ΑΒ ἴσον ἐστὶ τῷ ΣΤ, τουτέστι τῷ ἀπὸ τῆς ΛΝ τετραγώνῳ· ἡ ΛΝ ἄρα δύναται τὸ ΑΒ χωρίον.
And AK is also equal to LM, NX; therefore the remainder AB is equal to ST, that is, the square on LN; therefore LN produces the area AB.
λέγω, ὅτι ἡ ΛΝ μέσης ἀποτομή ἐστι δευτέρα.
I say that LN is a second apotome of a medial straight line.
ἐπεὶ γὰρ μέσα ἐδείχθη τὰ ΑΙ, ΖΚ καί ἐστιν ἴσα τοῖς ἀπὸ τῶν ΛΟ, ΟΝ, μέσον ἄρα καὶ ἑκάτερον τῶν ἀπὸ τῶν ΛΟ, ΟΝ· μέση ἄρα ἑκατέρα τῶν ΛΟ, ΟΝ. καὶ ἐπεὶ σύμμετρόν ἐστι τὸ ΑΙ τῷ ΖΚ, σύμμετρον ἄρα καὶ τὸ ἀπὸ τῆς ΛΟ τῷ ἀπὸ τῆς ΟΝ. πάλιν, ἐπεὶ ἀσύμμετρον ἐδείχθη τὸ ΑΙ τῷ ΕΚ, ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ΛΜ τῷ ΜΝ, τουτέστι τὸ ἀπὸ τῆς ΛΟ τῷ ὑπὸ τῶν ΛΟ, ΟΝ· ὥστε καὶ ἡ ΛΟ ἀσύμμετρός ἐστι τῇ ΟΝ· αἱ ΛΟ, ΟΝ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι.
For since the areas AI, ZK were proved medial and are equal to the squares on LO, ON, therefore each of the squares on LO, ON is also medial; therefore each of LO, ON is a medial straight line. And since AI is commensurable with ZK, therefore the square on LO is also commensurable with the square on ON. Again, since AI was proved incommensurable with EK, therefore LM is also incommensurable with MN, that is, the square on LO with the rectangle contained by LO, ON; so that LO is also incommensurable with ON; therefore LO, ON are medial straight lines commensurable in square only.
λέγω δή, ὅτι καὶ μέσον περιέχουσιν.
I say then that they also contain a medial area.
ἐπεὶ γὰρ μέσον ἐδείχθη τὸ ΕΚ καί ἐστιν ἴσον τῷ ὑπὸ τῶν ΛΟ, ΟΝ, μέσον ἄρα ἐστὶ καὶ τὸ ὑπὸ τῶν ΛΟ, ΟΝ· ὥστε αἱ ΛΟ, ΟΝ μέσαι εἰσὶ δυνάμει μόνον σύμμετροι μέσον περιέχουσαι.
For since EK was proved medial and is equal to the rectangle contained by LO, ON, therefore the rectangle contained by LO, ON is also medial; so that LO, ON are medial straight lines commensurable in square only containing a medial area.
ἡ ΛΝ ἄρα μέσης ἀποτομή ἐστι δευτέρα· καὶ δύναται τὸ ΑΒ χωρίον.
Therefore LN is a second apotome of a medial straight line; and it produces the area AB.
ἡ ἄρα τὸ ΑΒ χωρίον δυναμένη μέσης ἀποτομή ἐστι δευτέρα· ὅπερ ἔδει δεῖξαι.
Therefore the straight line producing the area AB is a second apotome of a medial straight line; which was to be proved.