For, since AD is a fourth binomial straight line, therefore AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line incommensurable with AE, and AE is commensurable in length with AB. []
ὁμοίως δὴ τοῖς προδεδειγμένοις δείξομεν, ὅτι ἡ ΜΞ ἐστιν ἡ τὸ ΑΓ χωρίον δυναμένη, καὶ αἱ ΜΝ, ΝΞ μέσαι εἰσὶ δυνάμει μόνον σύμμετροι· ὥστε ἡ ΜΞ ἐκ δύο μέσων ἐστίν. δεικτέον δή, ὅτι καὶ δευτέρα.
Let DE be bisected at F, and let there be applied to AE a parallelogram, the rectangle AG, GE, equal to the square on EF; therefore AG is incommensurable in length with GE. []
[Καὶ] ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΔΕ τῇ ΑΒ μήκει, τουτέστι τῇ ΕΚ, σύμμετρος δὲ ἡ ΔΕ τῇ ΕΖ, ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΖ τῇ ΕΚ μήκει. καί εἰσι ῥηταί·
Let GH, EK, FL be drawn parallel to AB, and let the rest of the construction be as before; it is then manifest that MO is thesideof the area AC.
αἱ ΖΕ, ΕΚ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. μέσον ἄρα [ἐστὶ] τὸ ΕΛ, τουτέστι τὸ ΜΡ· καὶ περιέχεται ὑπὸ τῶν ΜΝΞ·
It is next to be proved that MO is the irrational straight line called major.
μέσον ἄρα ἐστὶ τὸ ὑπὸ τῶν ΜΝΞ. ἡ ΜΞ ἄρα ἐκ δύο μέσων ἐστὶ δευτέρα·
Since AG is incommensurable with EG, AH is also incommensurable with GK, that is, SN with NQ; [, ] therefore MN, NO are incommensurable in square.
ὅπερ ἔδει δεῖξαι. ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων τετάρτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη μείζων.
And, since AE is commensurable with AB, AK is rational; [] and it is equal to the squares on MN, NO; therefore the sum of the squares on MN, NO is also rational.
χωρίον γὰρ τὸ ΑΓ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΒ καὶ τῆς ἐκ δύο ὀνομάτων τετάρτης τῆς ΑΔ διῃρημένης εἰς τὰ ὀνόματα κατὰ τὸ Ε, ὧν μεῖζον ἔστω τὸ ΑΕ·
And, since DE is incommensurable in length with AB, that is, with EK, while DE is commensurable with EF, therefore EF is incommensurable in length with EK. []
λέγω, ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη μείζων.
Therefore EK, EF are rational straight lines commensurable in square only; therefore LE, that is, MR, is medial. [] And it is contained by MN, NO; therefore the rectangle MN, NO is medial.
ἐπεὶ γὰρ ἡ ΑΔ ἐκ δύο ὀνομάτων ἐστὶ τετάρτη, αἱ ΑΕ, ΕΔ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ ἡ ΑΕ τῆς ΕΔ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ ἡ ΑΕ τῇ ΑΒ σύμμετρός [ἐστι] μήκει.
And the [sum] of the squares on MN, NO is rational, and MN, NO are incommensurable in square.
τετμήσθω ἡ ΔΕ δίχα κατὰ τὸ Ζ, καὶ τῷ ἀπὸ τῆς ΕΖ ἴσον παρὰ τὴν ΑΕ παραβεβλήσθω παραλληλόγραμμον τὸ ὑπὸ ΑΗ, ΗΕ·
But, if two straight lines incommensurable in square and making the sum of the squares on them rational, but the rectangle contained by them medial, be added together, the whole is irrational and is called major. []
ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΗ τῇ ΗΕ μήκει. ἤχθωσαν παράλληλοι τῇ ΑΒ αἱ ΗΘ, ΕΚ, ΖΛ, καὶ τὰ λοιπὰ τὰ αὐτὰ τοῖς πρὸ τούτου γεγονέτω·
Therefore MO is the irrational straight line called major and is thesideof the area AC. Q. E. D.
φανερὸν δή, ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἐστὶν ἡ ΜΞ. δεικτέον δή, ὅτι ἡ ΜΞ ἄλογός ἐστιν ἡ καλουμένη μείζων.
If an area be contained by a rational straight line and the fifth binomial, thesideof the area is the irrational straight line called the side of a rational plus a medial area.
ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΗ τῇ ΕΗ μήκει, ἀσύμμετρόν ἐστι καὶ τὸ ΑΘ τῷ ΗΚ, τουτέστι τὸ ΣΝ τῷ ΝΠ· αἱ ΜΝ, ΝΞ ἄρα δυνάμει εἰσὶν ἀσύμμετροι.
For let the area AC be contained by the rational straight line AB and the fifth binomial AD divided into its terms at E, so that AE is the greater term; I say that thesideof the area AC is the irrational straight line called the side of a rational plus a medial area.
καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΑΕ τῇ ΑΒ μήκει, ῥητόν ἐστι τὸ ΑΚ· καί ἐστιν ἴσον τοῖς ἀπὸ τῶν ΜΝ, ΝΞ· ῥητὸν ἄρα [ἐστὶ] καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ. καὶ ἐπεὶ ἀσύμμετρός [ἐστιν] ἡ ΔΕ τῇ ΑΒ μήκει, τουτέστι τῇ ΕΚ, ἀλλὰ ἡ ΔΕ σύμμετρός ἐστι τῇ ΕΖ, ἀσύμμετρος ἄρα ἡ ΕΖ τῇ ΕΚ μήκει.
For let the same construction be made as before shown; it is then manifest that MO is thesideof the area AC.
αἱ ΕΚ, ΕΖ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι·
It is then to be proved that MO is the side of a rational plus a medial area.
μέσον ἄρα τὸ ΛΕ, τουτέστι τὸ ΜΡ.
For, since AG is incommensurable with GE, [] therefore AH is also commensurable with HE, [, ] that is, the square on MN with the square on NO; therefore MN, NO are incommensurable in square.
καὶ περιέχεται ὑπὸ τῶν ΜΝ, ΝΞ· μέσον ἄρα ἐστὶ τὸ ὑπὸ τῶν ΜΝ, ΝΞ. καὶ ῥητὸν τὸ [συγκείμενον] ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ, καί εἰσιν ἀσύμμετροι αἱ ΜΝ, ΝΞ δυνάμει.
And, since AD is a fifth binomial straight line, and ED the lesser segment, therefore ED is commensurable in length with AB. [] But AE is incommensurable with ED; therefore AB is also incommensurable in length with AE. []
ἐὰν δὲ δύο εὐθεῖαι δυνάμει ἀσύμμετροι συντεθῶσι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων ῥητόν, τὸ δʼ ὑπʼ αὐτῶν μέσον, ἡ ὅλη ἄλογός ἐστιν, καλεῖται δὲ μείζων.
Therefore AK, that is, the sum of the squares on MN, NO, is medial. []
ἡ ΜΞ ἄρα ἄλογός ἐστιν ἡ καλουμένη μείζων, καὶ δύναται τὸ ΑΓ χωρίον· ὅπερ ἔδει δεῖξαι.
And, since DE is commensurable in length with AB, that is, with EK, while DE is commensurable with EF, therefore EF is also commensurable with EK. [] And EK is rational; therefore EL, that is, MR, that is, the rectangle MN, NO, is also rational. []
ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων πέμπτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη ῥητὸν καὶ μέσον δυναμένη.
Therefore MN, NO are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational.
χωρίον γὰρ τὸ ΑΓ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΒ καὶ τῆς ἐκ δύο ὀνομάτων πέμπτης τῆς ΑΔ διῃρημένης εἰς τὰ ὀνόματα κατὰ τὸ Ε, ὥστε τὸ μεῖζον ὄνομα εἶναι τὸ ΑΕ·
Therefore MO is the side of a rational plus a medial area [] and is thesideof the area AC. Q. E. D.
λέγω [δή], ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη ῥητὸν καὶ μέσον δυναμένη. Κατεσκευάσθω γὰρ τὰ αὐτὰ τοῖς πρότερον δεδειγμένοις·
If an area be contained by a rational straight line and the sixth binomial, thesideof the area is the irrational straight line called the side of the sum of two medial areas.
φανερὸν δή, ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἐστὶν ἡ ΜΞ. δεικτέον δή, ὅτι ἡ ΜΞ ἐστιν ἡ ῥητὸν καὶ μέσον δυναμένη.
For let the area ABCD be contained by the rational straight line AB and the sixth binomial AD, divided into its terms at E, so that AE is the greater term; I say that thesideof AC is the side of the sum of two medial areas.
ἐπεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΑΗ τῇ ΗΕ, ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ΑΘ τῷ ΘΕ, τουτέστι τὸ ἀπὸ τῆς ΜΝ τῷ ἀπὸ τῆς ΝΞ· αἱ ΜΝ, ΝΞ ἄρα δυνάμει εἰσὶν ἀσύμμετροι.
Let the same construction be made as before shown.
καὶ ἐπεὶ ἡ ΑΔ ἐκ δύο ὀνομάτων ἐστὶ πέμπτη, καί [ἐστιν] ἔλασσον αὐτῆς τμῆμα τὸ ΕΔ, σύμμετρος ἄρα ἡ ΕΔ τῇ ΑΒ μήκει.
It is then manifest that MO is thesideof AC, and that MN is incommensurable in square with NO.
ἀλλὰ ἡ ΑΕ τῇ ΕΔ ἐστιν ἀσύμμετρος·
Now, since EA is incommensurable in length with AB, therefore EA, AB are rational straight lines commensurable in square only; therefore AK, that is, the sum of the squares on MN, NO, is medial. []
καὶ ἡ ΑΒ ἄρα τῇ ΑΕ ἐστιν ἀσύμμετρος μήκει. [αἱ ΒΑ, ΑΕ ῥηταί εἰσι δυνάμει μόνον σύμμετροι.] μέσον ἄρα ἐστὶ τὸ ΑΚ, τουτέστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ.
Again, since ED is incommensurable in length with AB, therefore FE is also incommensurable with EK; [] therefore FE, EK are rational straight lines commensurable in square only; therefore EL, that is, MR, that is, the rectangle MN, NO, is medial. []
καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΔΕ τῇ ΑΒ μήκει, τουτέστι τῇ ΕΚ, ἀλλὰ ἡ ΔΕ τῇ ΕΖ σύμμετρός ἐστιν, καὶ ἡ ΕΖ ἄρα τῇ ΕΚ σύμμετρός ἐστιν. καὶ ῥητὴ ἡ ΕΚ· ῥητὸν ἄρα καὶ τὸ ΕΛ, τουτέστι τὸ ΜΡ, τουτέστι τὸ ὑπὸ ΜΝΞ·
And, since AE is incommensurable with EF, AK is also incommensurable with EL. [, ]
αἱ ΜΝ, ΝΞ ἄρα δυνάμει ἀσύμμετροί εἰσι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων μέσον, τὸ δʼ ὑπʼ αὐτῶν ῥητόν.
But AK is the sum of the squares on MN, NO, and EL is the rectangle MN, NO; therefore the sum of the squares on MN, NO is incommensurable with the rectangle MN, NO.
ἡ ΜΞ ἄρα ῥητὸν καὶ μέσον δυναμένη ἐστὶ καὶ δύναται τὸ ΑΓ χωρίον· ὅπερ ἔδει δεῖξαι.
And each of them is medial, and MN, NO are incommensurable in square. Therefore MO is the side of the sum of two medial areas [], and is thesideof AC. Q. E. D.
ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων ἕκτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη δύο μέσα δυναμένη.
[LEMMA. If a straight line be cut into unequal parts, the squares on the unequal parts are greater than twice the rectangle contained by the unequal parts.
χωρίον γὰρ τὸ ΑΒΓΔ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΒ καὶ τῆς ἐκ δύο ὀνομάτων ἕκτης τῆς ΑΔ διῃρημένης εἰς τὰ ὀνόματα κατὰ τὸ Ε, ὥστε τὸ μεῖζον ὄνομα εἶναι τὸ ΑΕ·
Let AB be a straight line, and let it be cut into unequal parts at C, and let AC be the greater; I say that the squares on AC, CB are greater than twice the rectangle AC, CB.
λέγω, ὅτι ἡ τὸ ΑΓ δυναμένη ἡ δύο μέσα δυναμένη ἐστίν. Κατεσκευάσθω [γὰρ] τὰ αὐτὰ τοῖς προδεδειγμένοις.
For let AB be bisected at D.
φανερὸν δή, ὅτι [ἡ] τὸ ΑΓ δυναμένη ἐστὶν ἡ ΜΞ, καὶ ὅτι ἀσύμμετρός ἐστι ἡ ΜΝ τῇ ΝΞ δυνάμει.
Since then a straight line has been cut into equal parts at D, and into unequal parts at C, therefore the rectangle AC, CB together with the square on CD is equal to the square on AD, [] so that the rectangle AC, CB is less than double of the square on AD.
καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΕΑ τῇ ΑΒ μήκει, αἱ ΕΑ, ΑΒ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· μέσον ἄρα ἐστὶ τὸ ΑΚ, τουτέστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ.
But the squares on AC, CB are double of the squares on AD, DC; [] therefore the squares on AC, CB are greater than twice the rectangle AC, CB. Q. E. D.]
πάλιν, ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΕΔ τῇ ΑΒ μήκει, ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΖΕ τῇ ΕΚ· αἱ ΖΕ, ΕΚ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι·
The square on the binomial straight line applied to a rational straight line produces as breadth the first binomial.
μέσον ἄρα ἐστὶ τὸ ΕΛ, τουτέστι τὸ ΜΡ, τουτέστι τὸ ὑπὸ τῶν ΜΝΞ. καὶ ἐπεὶ ἀσύμμετρος ἡ ΑΕ τῇ ΕΖ, καὶ τὸ ΑΚ τῷ ΕΛ ἀσύμμετρόν ἐστιν.
Let AB be a binomial straight line divided into its terms at C, so that AC is the greater term; let a rational straight line DE be set out, and let DEFG equal to the square on AB be applied to DE producing DG as its breadth; I say that DG is a first binomial straight line.
ἀλλὰ τὸ μὲν ΑΚ ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ, τὸ δὲ ΕΛ ἐστι τὸ ὑπὸ τῶν ΜΝΞ· ἀσύμμετρον ἄρα ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝΞ τῷ ὑπὸ τῶν ΜΝΞ. καί ἐστι μέσον ἑκάτερον αὐτῶν, καὶ αἱ ΜΝ, ΝΞ δυνάμει εἰσὶν ἀσύμμετροι.
For let there be applied to DE the rectangle DH equal to the square on AC, and KL equal to the square on BC; therefore the remainder, twice the rectangle AC, CB, is equal to MF.
ἡ ΜΞ ἄρα δύο μέσα δυναμένη ἐστὶ καὶ δύναται τὸ ΑΓ· ὅπερ ἔδει δεῖξαι.
Let MG be bisected at N, and let NO be drawn parallel [to ML or GF].
[λῆμμα] [ἐὰν εὐθεῖα γραμμὴ τμηθῇ εἰς ἄνισα, τὰ ἀπὸ τῶν ἀνίσων τετράγωνα μείζονά ἐστι τοῦ δὶς ὑπὸ τῶν ἀνίσων περιεχομένου ὀρθογωνίου.
Therefore each of the rectangles MO, NF is equal to once the rectangle AC, CB.
] [ἔστω εὐθεῖα ἡ ΑΒ καὶ τετμήσθω εἰς ἄνισα κατὰ τὸ Γ, καὶ ἔστω μείζων ἡ ΑΓ·
Now, since AB is a binomial divided into its terms at C, therefore AC, CB are rational straight lines commensurable in square only; [] therefore the squares on AC, CB are rational and commensurable with one another, so that the sum of the squares on AC, CB is also rational. [] And it is equal to DL; therefore DL is rational.
λέγω, ὅτι τὰ ἀπὸ τῶν ΑΓ, ΓΒ μείζονά ἐστι τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ. ] [τετμήσθω γὰρ ἡ ΑΒ δίχα κατὰ τὸ Δ. ἐπεὶ] [οὖν εὐθεῖα γραμμὴ τέτμηται εἰς μὲν ἴσα κατὰ τὸ Δ, εἰς δὲ ἄνισα κατὰ τὸ Γ, τὸ ἄρα ὑπὸ τῶν ΑΓ, ΓΒ μετὰ τοῦ ἀπὸ ΓΔ ἴσον ἐστὶ τῷ ἀπὸ ΑΔ·
And it is applied to the rational straight line DE; therefore DM is rational and commensurable in length with DE. []
ὥστε τὸ ὑπὸ τῶν ΑΓ, ΓΒ ἔλαττόν ἐστι τοῦ ἀπὸ ΑΔ· τὸ ἄρα δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἔλαττον ἢ διπλάσιόν ἐστι τοῦ ἀπὸ ΑΔ.
Again, since AC, CB are rational straight lines commensurable in square only, therefore twice the rectangle AC, CB, that is MF, is medial. []
ἀλλὰ τὰ ἀπὸ τῶν ΑΓ, ΓΒ διπλάσιά [ἐστι] τῶν ἀπὸ τῶν ΑΔ, ΔΓ· τὰ ἄρα ἀπὸ τῶν ΑΓ, ΓΒ μείζονά ἐστι τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ·
And it is applied to the rational straight line ML; therefore MG is also rational and incommensurable in length with ML, that is, DE. [] But MD is also rational and is commensurable in length with DE; therefore DM is incommensurable in length with MG. []
ὅπερ ἔδει δεῖξαι.] τὸ ἀπὸ τῆς ἐκ δύο ὀνομάτων παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων πρώτην.
And they are rational; therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. []
ἔστω ἐκ δύο ὀνομάτων ἡ ΑΒ διῃρημένη εἰς τὰ ὀνόματα κατὰ τὸ Γ, ὥστε τὸ μεῖζον ὄνομα εἶναι τὸ ΑΓ, καὶ ἐκκείσθω ῥητὴ ἡ ΔΕ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΔΕ παραβεβλήσθω τὸ ΔΕΖΗ πλάτος ποιοῦν τὴν ΔΗ·
It is next to be proved that it is also a first binomial straight line.
λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ πρώτη.
Since the rectangle AC, CB is a mean proportional between the squares on AC, CB, [cf. ] therefore MO is also a mean proportional between DH, KL.
παραβεβλήσθω γὰρ παρὰ τὴν ΔΕ τῷ μὲν ἀπὸ τῆς ΑΓ ἴσον τὸ ΔΘ, τῷ δὲ ἀπὸ τῆς ΒΓ ἴσον τὸ ΚΛ· λοιπὸν ἄρα τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἴσον ἐστὶ τῷ ΜΖ.
Therefore, as DH is to MO, so is MO to KL, that is, as DK is to MN, so is MN to MK; [] therefore the rectangle DK, KM is equal to the square on MN. []
τετμήσθω ἡ ΜΗ δίχα κατὰ τὸ Ν, καὶ παράλληλος ἤχθω ἡ ΝΞ [ἑκατέρᾳ τῶν ΜΛ, ΗΖ ].
And, since the square on AC is commensurable with the square on CB, DH is also commensurable with KL, so that DK is also commensurable with KM. [, ]
ἑκάτερον ἄρα τῶν ΜΞ, ΝΖ ἴσον ἐστὶ τῷ ἅπαξ ὑπὸ τῶν ΑΓΒ. καὶ ἐπεὶ ἐκ δύο ὀνομάτων ἐστὶν ἡ ΑΒ διῃρημένη εἰς τὰ ὀνόματα κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι·
And, since the squares on AC, CB are greater than twice the rectangle AC, CB, [] therefore DL is also greater than MF, so that DM is also greater than MG. []
τὰ ἄρα ἀπὸ τῶν ΑΓ, ΓΒ ῥητά ἐστι καὶ σύμμετρα ἀλλήλοις· ὥστε καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ [σύμμετρόν ἐστι τοῖς ἀπὸ τῶν ΑΓ, ΓΒ·
And the rectangle DK, KM is equal to the square on MN, that is, to the fourth part of the square on MG, and DK is commensurable with KM.
ῥητὸν ἄρα ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ ]. καί ἐστιν ἴσον τῷ ΔΛ·
But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into commensurable parts, the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater; [] therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM.
ῥητὸν ἄρα ἐστὶ τὸ ΔΛ. καὶ παρὰ ῥητὴν τὴν ΔΕ παράκειται· ῥητὴ ἄρα ἐστὶν ἡ ΔΜ καὶ σύμμετρος τῇ ΔΕ μήκει. πάλιν, ἐπεὶ αἱ ΑΓ, ΓΒ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, μέσον ἄρα ἐστὶ τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, τουτέστι τὸ ΜΖ. καὶ παρὰ ῥητὴν τὴν ΜΛ παράκειται· ῥητὴ ἄρα καὶ ἡ ΜΗ ἐστι καὶ ἀσύμμετρος τῇ ΜΛ, τουτέστι τῇ ΔΕ, μήκει. ἔστι δὲ καὶ ἡ ΜΔ ῥητὴ καὶ τῇ ΔΕ μήκει σύμμετρος· ἀσύμμετρος ἄρα ἐστὶν ἡ ΔΜ τῇ ΜΗ μήκει.
And DM, MG are rational, and DM, which is the greater term, is commensurable in length with the rational straight line DE set out.
καί εἰσι ῥηταί· αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ.
Therefore DG is a first binomial straight line. [] Q. E. D.
δεικτέον δή, ὅτι καὶ πρώτη.
The square on the first bimedial straight line applied to a rational straight line produces as breadth the second binomial.
ἐπεὶ τῶν ἀπὸ τῶν ΑΓ, ΓΒ μέσον ἀνάλογόν ἐστι τὸ ὑπὸ τῶν ΑΓΒ, καὶ τῶν ΔΘ, ΚΛ ἄρα μέσον ἀνάλογόν ἐστι τὸ ΜΞ.
Let AB be a first bimedial straight line divided into its medials at C, of which medials AC is the greater; let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a second binominal straight line. For let the same construction as before be made.
ἔστιν ἄρα ὡς τὸ ΔΘ πρὸς τὸ ΜΞ, οὕτως τὸ ΜΞ πρὸς τὸ ΚΛ, τουτέστιν ὡς ἡ ΔΚ πρὸς τὴν ΜΝ, ἡ ΜΝ πρὸς τὴν ΜΚ· τὸ ἄρα ὑπὸ τῶν ΔΚ, ΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΜΝ. καὶ ἐπεὶ σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΓ τῷ ἀπὸ τῆς ΓΒ, σύμμετρόν ἐστι καὶ τὸ ΔΘ τῷ ΚΛ· ὥστε καὶ ἡ ΔΚ τῇ ΚΜ σύμμετρός ἐστιν.
Then, since AB is a first bimedial divided at C, therefore AC, CB are medial straight lines commensurable in square only, and containing a rational rectangle, [] so that the squares on AC, CB are also medial. [] Therefore DL is medial. [] And it has been applied to the rational straight line DE; therefore MD is rational and incommensurable in length with DE. []
καὶ ἐπεὶ μείζονά ἐστι τὰ ἀπὸ τῶν ΑΓ, ΓΒ τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, μεῖζον ἄρα καὶ τὸ ΔΛ τοῦ ΜΖ· ὥστε καὶ ἡ ΔΜ τῆς ΜΗ μείζων ἐστίν. καί ἐστιν ἴσον τὸ ὑπὸ τῶν ΔΚ, ΚΜ τῷ ἀπὸ τῆς ΜΝ, τουτέστι τῷ τετάρτῳ τοῦ ἀπὸ τῆς ΜΗ, καὶ σύμμετρος ἡ ΔΚ τῇ ΚΜ.
Again, since twice the rectangle AC, CB is rational, MF is also rational.
ἐὰν δὲ ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ καὶ εἰς σύμμετρα αὐτὴν διαιρῇ, ἡ μείζων τῆς ἐλάσσονος μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ·
And it is applied to the rational straight line ML; therefore MG is also rational and commensurable in length with ML, that is, DE; [] therefore DM is incommensurable in length with MG. []
ἡ ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ.
And they are rational; therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. []
καί εἰσι ῥηταὶ αἱ ΔΜ, ΜΗ, καὶ ἡ ΔΜ μεῖζον ὄνομα σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΔΕ μήκει. ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ πρώτη·
It is next to be proved that it is also a second binomial straight line.
ὅπερ ἔδει δεῖξαι.
For, since the squares on AC, CB are greater than twice the rectangle AC, CB, therefore DL is also greater than MF, so that DM is also greater than MG. []
τὸ ἀπὸ τῆς ἐκ δύο μέσων πρώτης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων δευτέραν.
And, since the square on AC is commensurable with the square on CB, DH is also commensurable with KL, so that DK is also commensurable with KM. [, ]
ἔστω ἐκ δύο μέσων πρώτη ἡ ΑΒ διῃρημένη εἰς τὰς μέσας κατὰ τὸ Γ, ὧν μείζων ἡ ΑΓ, καὶ ἐκκείσθω ῥητὴ ἡ ΔΕ, καὶ παραβεβλήσθω παρὰ τὴν ΔΕ τῷ ἀπὸ τῆς ΑΒ ἴσον παραλληλόγραμμον τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ·
And the rectangle DK, KM is equal to the square on MN; therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM. []
λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ δευτέρα. Κατεσκευάσθω γὰρ τὰ αὐτὰ τοῖς πρὸ τούτου.
And MG is commensurable is length with DE. Therefore DG is a second binomial straight line. []
καὶ ἐπεὶ ἡ ΑΒ ἐκ δύο μέσων ἐστὶ πρώτη διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι ῥητὸν περιέχουσαι·
The square on the second bimedial straight line applied to a rational straight line produces as breadth the third binomial.
ὥστε καὶ τὰ ἀπὸ τῶν ΑΓ, ΓΒ μέσα ἐστίν. μέσον ἄρα ἐστὶ τὸ ΔΛ. καὶ παρὰ ῥητὴν τὴν ΔΕ παραβέβληται·
Let AB be a second bimedial straight line divided into its medials at C, so that AC is the greater segment; let DE be any rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth; I say that DG is a third binomial straight line.
ῥητὴ ἄρα ἐστίν ἡ ΜΔ καὶ ἀσύμμετρος τῇ ΔΕ μήκει. πάλιν, ἐπεὶ ῥητόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, ῥητόν ἐστι καὶ τὸ ΜΖ. καὶ παρὰ ῥητὴν τὴν ΜΛ παράκειται· ῥητὴ ἄρα [ἐστὶ] καὶ ἡ ΜΗ καὶ μήκει σύμμετρος τῇ ΜΛ, τουτέστι τῇ ΔΕ· ἀσύμμετρος ἄρα ἐστὶν ἡ ΔΜ τῇ ΜΗ μήκει.
Let the same construction be made as before shown.
καί εἰσι ῥηταί· αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι·
Then, since AB is a second bimedial divided at C, therefore AC, CB are medial straight lines commensurable in square only and containing a medial rectangle, [] so that the sum of the squares on AC, CB is also medial. []
ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ. δεικτέον δή, ὅτι καὶ δευτέρα. ἐπεὶ γὰρ τὰ ἀπὸ τῶν ΑΓ, ΓΒ μείζονά ἐστι τοῦ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, μεῖζον ἄρα καὶ τὸ ΔΛ τοῦ ΜΖ·
And it is equal to DL; therefore DL is also medial.
ὥστε καὶ ἡ ΔΜ τῆς ΜΗ.
And it is applied to the rational straight line DE; therefore MD is also rational and incommensurable in length with DE. []
καὶ ἐπεὶ σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΓ τῷ ἀπὸ τῆς ΓΒ, σύμμετρόν ἐστι καὶ τὸ ΔΘ τῷ ΚΛ· ὥστε καὶ ἡ ΔΚ τῇ ΚΜ σύμμετρός ἐστιν.
For the same reason, MG is also rational and incommensurable in length with ML, that is, with DE; therefore each of the straight lines DM, MG is rational and incommensurable in length with DE.
καί ἐστι τὸ ὑπὸ τῶν ΔΚΜ ἴσον τῷ ἀπὸ τῆς ΜΝ· ἡ ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. καί ἐστιν ἡ ΜΗ σύμμετρος τῇ ΔΕ μήκει. ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ δευτέρα.
And, since AC is incommensurable in length with CB, and, as AC is to CB, so is the square on AC to the rectangle AC, CB, therefore the square on AC is also incommensurable with the rectangle AC, CB. []
τὸ ἀπὸ τῆς ἐκ δύο μέσων δευτέρας παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων τρίτην.
Hence the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, [] that is, DL is incommensurable with MF, so that DM is also incommensurable with MG. [, ]
ἔστω ἐκ δύο μέσων δευτέρα ἡ ΑΒ διῃρημένη εἰς τὰς μέσας κατὰ τὸ Γ, ὥστε τὸ μεῖζον τμῆμα εἶναι τὸ ΑΓ, ῥητὴ δέ τις ἔστω ἡ ΔΕ, καὶ παρὰ τὴν ΔΕ τῷ ἀπὸ τῆς ΑΒ ἴσον παραλληλόγραμμον παραβεβλήσθω τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ·
And they are rational; therefore DG is binomial. []
λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ τρίτη.
It is to be proved that it is also a third binomial straight line.
Κατεσκευάσθω τὰ αὐτὰ τοῖς προδεδειγμένοις.
In manner similar to the foregoing we may conclude that DM is greater than MG, and that DK is commensurable with KM.
καὶ ἐπεὶ ἐκ δύο μέσων δευτέρα ἐστὶν ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι μέσον περιέχουσαι·
And the rectangle DK, KM is equal to the square on MN; therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM.
ὥστε καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ μέσον ἐστίν. καί ἐστιν ἴσον τῷ ΔΛ· μέσον ἄρα καὶ τὸ ΔΛ.
And neither of the straight lines DM, MG is commensurable in length with DE. Therefore DG is a third binomial straight line. [] Q. E. D.
καὶ παράκειται παρὰ ῥητὴν τὴν ΔΕ· ῥητὴ ἄρα ἐστὶ καὶ ἡ ΜΔ καὶ ἀσύμμετρος τῇ ΔΕ μήκει.
The square on the major straight line applied to a rational straight line produces as breadth the fourth binomial.
διὰ τὰ αὐτὰ δὴ καὶ ἡ ΜΗ ῥητή ἐστι καὶ ἀσύμμετρος τῇ ΜΛ, τουτέστι τῇ ΔΕ, μήκει· ῥητὴ ἄρα ἐστὶν ἑκατέρα τῶν ΔΜ, ΜΗ καὶ ἀσύμμετρος τῇ ΔΕ μήκει.
Let AB be a major straight line divided at C, so that AC is greater than CB; let DE be a rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth; I say that DG is a fourth binomial straight line. Let the same construction be made as before shown.
καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΓ τῇ ΓΒ μήκει, ὡς δὲ ἡ ΑΓ πρὸς τὴν ΓΒ, οὕτως τὸ ἀπὸ τῆς ΑΓ πρὸς τὸ ὑπὸ τῶν ΑΓΒ, ἀσύμμετρον ἄρα καὶ τὸ ἀπὸ τῆς ΑΓ τῷ ὑπὸ τῶν ΑΓΒ. ὥστε καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ τῷ δὶς ὑπὸ τῶν ΑΓΒ ἀσύμμετρόν ἐστιν, τουτέστι τὸ ΔΛ τῷ ΜΖ·
Then, since AB is a major straight line divided at C, AC, CB are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. []
ὥστε καὶ ἡ ΔΜ τῇ ΜΗ ἀσύμμετρός ἐστιν. καί εἰσι ῥηταί· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ. δεικτέον [δή], ὅτι καὶ τρίτη.
Since then the sum of the squares on AC, CB is rational, therefore DL is rational; therefore DM is also rational and commensurable in length with DE. []
ὁμοίως δὴ τοῖς προτέροις ἐπιλογιούμεθα, ὅτι μείζων ἐστὶν ἡ ΔΜ τῆς ΜΗ, καὶ σύμμετρος ἡ ΔΚ τῇ ΚΜ. καί ἐστι τὸ ὑπὸ τῶν ΔΚΜ ἴσον τῷ ἀπὸ τῆς ΜΝ· ἡ ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ.
Again, since twice the rectangle AC, CB, that is, MF, is medial, and it is applied to the rational straight line ML, therefore MG is also rational and incommensurable in length with DE; [] therefore DM is also incommensurable in length with MG. [] Therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. []
καὶ οὐδετέρα τῶν ΔΜ, ΜΗ σύμμετρός ἐστι τῇ ΔΕ μήκει. ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ τρίτη· ὅπερ ἔδει δεῖξαι. τὸ ἀπὸ τῆς μείζονος παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων τετάρτην.
It is to be proved that it is also a fourth binomial straight line.
ἔστω μείζων ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, ὥστε μείζονα εἶναι τὴν ΑΓ τῆς ΓΒ, ῥητὴ δὲ ἡ ΔΕ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΔΕ παραβεβλήσθω τὸ ΔΖ παραλληλόγραμμον πλάτος ποιοῦν τὴν ΔΗ·
In manner similar to the foregoing we can prove that DM is greater than MG, and that the rectangle DK, KM is equal to the square on MN.
λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ τετάρτη.
Since then the square on AC is incommensurable with the square on CB, therefore DH is also incommensurable with KL, so that DK is also incommensurable with KM. [, ]
Κατεσκευάσθω τὰ αὐτὰ τοῖς προδεδειγμένοις. καὶ ἐπεὶ μείζων ἐστὶν ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων ῥητόν, τὸ δὲ ὑπʼ αὐτῶν μέσον.
But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into incommensurable parts, then the square on the greater will be greater than the square on the less by the square on a straight line incommensurable in length with the greater; [] therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM.
ἐπεὶ οὖν ῥητόν ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, ῥητὸν ἄρα ἐστὶ τὸ ΔΛ· ῥητὴ ἄρα καὶ ἡ ΔΜ καὶ σύμμετρος τῇ ΔΕ μήκει. πάλιν, ἐπεὶ μέσον ἐστὶ τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, τουτέστι τὸ ΜΖ, καὶ παρὰ ῥητήν ἐστι τὴν ΜΛ, ῥητὴ ἄρα ἐστὶ καὶ ἡ ΜΗ καὶ ἀσύμμετρος τῇ ΔΕ μήκει· ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΔΜ τῇ ΜΗ μήκει. αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι·
And DM, MG are rational straight lines commensurable in square only, and DM is commensurable with the rational straight line DE set out.
ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ. δεικτέον [δή], ὅτι καὶ τετάρτη.
Therefore DG is a fourth binomial straight line. [] Q. E. D.
ὁμοίως δὴ δείξομεν τοῖς πρότερον, ὅτι μείζων ἐστὶν ἡ ΔΜ τῆς ΜΗ, καὶ ὅτι τὸ ὑπὸ ΔΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΜΝ.
The square on the side of a rational plus a medial area applied to a rational straight line produces as breadth the fifth binomial.
ἐπεὶ οὖν ἀσύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΓ τῷ ἀπὸ τῆς ΓΒ, ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ΔΘ τῷ ΚΛ·
Let AB be the side of a rational plus a medial area, divided into its straight lines at C, so that AC is the greater; let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a fifth binomial straight line. Let the same construction as before be made.
ὥστε ἀσύμμετρος καὶ ἡ ΔΚ τῇ ΚΜ ἐστιν. ἐὰν δὲ ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παραλληλόγραμμον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ καὶ εἰς ἀσύμμετρα αὐτὴν διαιρῇ, ἡ μείζων τῆς ἐλάσσονος μεῖζον δυνήσεται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει·
Since then AB is the side of a rational plus a medial area, divided at C, therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. []
ἡ ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί εἰσιν αἱ ΔΜ, ΜΗ ῥηταὶ δυνάμει μόνον σύμμετροι, καὶ ἡ ΔΜ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΔΕ. ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ τετάρτη·
Since then the sum of the squares on AC, CB is medial, therefore DL is medial, so that DM is rational and incommensurable in length with DE. []
ὅπερ ἔδει δεῖξαι. τὸ ἀπὸ τῆς ῥητὸν καὶ μέσον δυναμένης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων πέμπτην.
Again, since twice the rectangle AC, CB, that is MF, is rational, therefore MG is rational and commensurable with DE. []
ἔστω ῥητὸν καὶ μέσον δυναμένη ἡ ΑΒ διῃρημένη εἰς τὰς εὐθείας κατὰ τὸ Γ, ὥστε μείζονα εἶναι τὴν ΑΓ, καὶ ἐκκείσθω ῥητὴ ἡ ΔΕ, καὶ τῷ ἀπὸ τῆς ΑΒ ἴσον παρὰ τὴν ΔΕ παραβεβλήσθω τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ·
Therefore DM is incommensurable with MG; [] therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. []
λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶ πέμπτη.
I say next that it is also a fifth binomial straight line.
Κατεσκευάσθω τὰ αὐτὰ τοῖς πρὸ τούτου.
For it can be proved similarly that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM; therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM. []
ἐπεὶ οὖν ῥητὸν καὶ μέσον δυναμένη ἐστὶν ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων μέσον, τὸ δʼ ὑπʼ αὐτῶν ῥητόν.
And DM, MG are commensurable in square only, and the less, MG, is commensurable in length with DE.
ἐπεὶ οὖν μέσον ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ, μέσον ἄρα ἐστὶ τὸ ΔΛ·
Therefore DG is a fifth binomial. Q. E. D.
ὥστε ῥητή ἐστιν ἡ ΔΜ καὶ μήκει ἀσύμμετρος τῇ ΔΕ.
The square on the side of the sum of two medial areas applied to a rational straight line produces as breadth the sixth binomial.
πάλιν, ἐπεὶ ῥητόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΓΒ, τουτέστι τὸ ΜΖ, ῥητὴ ἄρα ἡ ΜΗ καὶ σύμμετρος τῇ ΔΕ.
Let AB be the side of the sum of two medial areas, divided at C, let DE be a rational straight line, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a sixth binomial straight line. For let the same construction be made as before.
ἀσύμμετρος ἄρα ἡ ΔΜ τῇ ΜΗ· αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ. λέγω δή, ὅτι καὶ πέμπτη. ὁμοίως γὰρ δειχθήσεται, ὅτι τὸ ὑπὸ τῶν ΔΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΜΝ, καὶ ἀσύμμετρος ἡ ΔΚ τῇ ΚΜ μήκει·
Then, since AB is the side of the sum of two medial areas, divided at C, therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and moreover the sum of the squares on them incommensurable with the rectangle contained by them, [] so that, in accordance with what was before proved, each of the rectangles DL, MF is medial.
ἡ ΔΜ ἄρα τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί εἰσιν αἱ ΔΜ, ΜΗ [ῥηταὶ] δυνάμει μόνον σύμμετροι, καὶ ἡ ἐλάσσων ἡ ΜΗ σύμμετρος τῇ ΔΕ μήκει. ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ πέμπτη· ὅπερ ἔδει δεῖξαι. τὸ ἀπὸ τῆς δύο μέσα δυναμένης παρὰ ῥητὴν παραβαλλόμενον πλάτος ποιεῖ τὴν ἐκ δύο ὀνομάτων ἕκτην. ἔστω δύο μέσα δυναμένη ἡ ΑΒ διῃρημένη κατὰ τὸ Γ, ῥητὴ δὲ ἔστω ἡ ΔΕ.
And they are applied to the rational straight line DE; therefore each of the straight lines DM, MG is rational and incommensurable in length with DE. []
καὶ παρὰ τὴν ΔΕ τῷ ἀπὸ τῆς ΑΒ ἴσον παραβεβλήσθω τὸ ΔΖ πλάτος ποιοῦν τὴν ΔΗ· λέγω, ὅτι ἡ ΔΗ ἐκ δύο ὀνομάτων ἐστὶν ἕκτη.
And, since the sum of the squares on AC, CB is incommensurable with twice the rectangle AC, CB, therefore DL is incommensurable with MF.
Κατεσκευάσθω γὰρ τὰ αὐτὰ τοῖς πρότερον.
Therefore DM is also incommensurable with MG; [, ] therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [] I say next that it is also a sixth binomial straight line.
καὶ ἐπεὶ ἡ ΑΒ δύο μέσα δυναμένη ἐστὶ διῃρημένη κατὰ τὸ Γ, αἱ ΑΓ, ΓΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων μέσον καὶ τὸ ὑπʼ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τὸ ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων συγκείμενον τῷ ὑπʼ αὐτῶν·
Similarly again we can prove that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM; and, for the same reason, the square on DM is greater than the square on MG by the square on a straight line incommensurable in length with DM.
ὥστε κατὰ τὰ προδεδειγμένα μέσον ἐστὶν ἑκάτερον τῶν ΔΛ, ΜΖ. καὶ παρὰ ῥητὴν τὴν ΔΕ παράκειται· ῥητὴ ἄρα ἐστὶν ἑκατέρα τῶν ΔΜ, ΜΗ καὶ ἀσύμμετρος τῇ ΔΕ μήκει. καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, ἀσύμμετρον ἄρα ἐστὶ τὸ ΔΛ τῷ ΜΖ.
And neither of the straight lines DM, MG is commensurable in length with the rational straight line DE set out.
ἀσύμμετρος ἄρα καὶ ἡ ΔΜ τῇ ΜΗ·
Therefore DG is a sixth binomial straight line. Q. E. D.
αἱ ΔΜ, ΜΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι·
A straight line commensurable in length with a binomial straight line is itself also binomial and the same in order.
ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΔΗ. λέγω δή, ὅτι καὶ ἕκτη.
Let AB be binomial, and let CD be commensurable in length with AB; I say that CD is binomial and the same in order with AB.
ὁμοίως δὴ πάλιν δείξομεν, ὅτι τὸ ὑπὸ τῶν ΔΚΜ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΜΝ, καὶ ὅτι ἡ ΔΚ τῇ ΚΜ μήκει ἐστὶν ἀσύμμετρος· καὶ διὰ τὰ αὐτὰ δὴ ἡ ΔΜ τῆς ΜΗ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει.
For, since AB is binomial, let it be divided into its terms at E, and let AE be the greater term; therefore AE, EB are rational straight lines commensurable in square only. []
καὶ οὐδετέρα τῶν ΔΜ, ΜΗ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΔΕ μήκει. ἡ ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶν ἕκτη·
Let it be contrived that, as AB is to CD, so is AE to CF; [] therefore also the remainder EB is to the remainder FD as AB is to CD. []
ὅπερ ἔδει δεῖξαι. ἡ τῇ ἐκ δύο ὀνομάτων μήκει σύμμετρος καὶ αὐτὴ ἐκ δύο ὀνομάτων ἐστὶ καὶ τῇ τάξει ἡ αὐτή.
But AB is commensurable in length with CD; therefore AE is also commensurable with CF, and EB with FD. []
ἔστω ἐκ δύο ὀνομάτων ἡ ΑΒ, καὶ τῇ ΑΒ μήκει σύμμετρος ἔστω ἡ ΓΔ·
And AE, EB are rational; therefore CF, FD are also rational. And, as AE is to CF, so is EB to FD. []
λέγω, ὅτι ἡ ΓΔ ἐκ δύο ὀνομάτων ἐστὶ καὶ τῇ τάξει ἡ αὐτὴ τῇ ΑΒ.
Therefore, alternately, as AE is to EB, so is CF to FD. []
ἐπεὶ γὰρ ἐκ δύο ὀνομάτων ἐστὶν ἡ ΑΒ, διῃρήσθω εἰς τὰ ὀνόματα κατὰ τὸ Ε, καὶ ἔστω μεῖζον ὄνομα τὸ ΑΕ·
But AE, EB are commensurable in square only; therefore CF, FD are also commensurable in square only. [] And they are rational; therefore CD is binomial. []
αἱ ΑΕ, ΕΒ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. γεγονέτω ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΑΕ πρὸς τὴν ΓΖ·
I say next that it is the same in order with AB.
καὶ λοιπὴ ἄρα ἡ ΕΒ πρὸς λοιπὴν τὴν ΖΔ ἐστιν, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ.
For the square on AE is greater than the square on EB either by the square on a straight line commensurable with AE or by the square on a straight line incommensurable with it.
σύμμετρος δὲ ἡ ΑΒ τῇ ΓΔ μήκει. σύμμετρος ἄρα ἐστὶ καὶ ἡ μὲν ΑΕ τῇ ΓΖ, ἡ δὲ ΕΒ τῇ ΖΔ. καί εἰσι ῥηταὶ αἱ ΑΕ, ΕΒ· ῥηταὶ ἄρα εἰσὶ καὶ αἱ ΓΖ, ΖΔ.
If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line commensurable with CF. []
καὶ [ἐπεί] ἐστιν ὡς ἡ ΑΕ πρὸς ΓΖ, ἡ ΕΒ πρὸς ΖΔ. ἐναλλὰξ ἄρα ἐστὶν ὡς ἡ ΑΕ πρὸς ΕΒ, ἡ ΓΖ πρὸς ΖΔ. αἱ δὲ ΑΕ, ΕΒ δυνάμει μόνον [εἰσὶ] σύμμετροι· καὶ αἱ ΓΖ, ΖΔ ἄρα δυνάμει μόνον εἰσὶ σύμμετροι.
And, if AE is commensurable with the rational straight line set out, CF will also be commensurable with it, [] and for this reason each of the straight lines AB, CD is a first binomial, that is, the same in order. []
καί εἰσι ῥηταί· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΓΔ. λέγω δή, ὅτι τῇ τάξει ἐστὶν ἡ αὐτὴ τῇ ΑΒ. ἡ γὰρ ΑΕ τῆς ΕΒ μεῖζον δύναται ἤτοι τῷ ἀπὸ συμμέτρου ἑαυτῇ ἢ τῷ ἀπὸ ἀσυμμέτρου.
But, if EB is commensurable with the rational straight line set out, FD is also commensurable with it, [] and for this reason again CD will be the same in order with AB, for each of them will be a second binomial. []
εἰ μὲν οὖν ἡ ΑΕ τῆς ΕΒ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καὶ ἡ ΓΖ τῆς ΖΔ μεῖζον δυνήσεται τῷ ἀπὸ συμμέτρου ἑαυτῇ.
But, if neither of the straight lines AE, EB is commensurable with the rational straight line set out, neither of the straight lines CF, FD will be commensurable with it, [] and each of the straight lines AB, CD is a third binomial. []
καὶ εἰ μὲν σύμμετρός ἐστιν ἡ ΑΕ τῇ ἐκκειμένῃ ῥητῇ, καὶ ἡ ΓΖ σύμμετρος αὐτῇ ἔσται, καὶ διὰ τοῦτο ἑκατέρα τῶν ΑΒ, ΓΔ ἐκ δύο ὀνομάτων ἐστὶ πρώτη, τουτέστι τῇ τάξει ἡ αὐτή. εἰ δὲ ἡ ΕΒ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ, καὶ ἡ ΖΔ σύμμετρός ἐστιν αὐτῇ, καὶ διὰ τοῦτο πάλιν τῇ τάξει ἡ αὐτὴ ἔσται τῇ ΑΒ·
But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, the square on CF is also greater than the square on FD by the square on a straight line incommensurable with CF. []
ἑκατέρα γὰρ αὐτῶν ἔσται ἐκ δύο ὀνομάτων δευτέρα. εἰ δὲ οὐδετέρα τῶν ΑΕ, ΕΒ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ, οὐδετέρα τῶν ΓΖ, ΖΔ σύμμετρος αὐτῇ ἔσται, καί ἐστιν ἑκατέρα τρίτη.
And, if AE is commensurable with the rational straight line set out, CF is also commensurable with it, and each of the straight lines AB, CD is a fourth binomial. []
εἰ δὲ ἡ ΑΕ τῆς ΕΒ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ ἡ ΓΖ τῆς ΖΔ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ.
But, if EB is so commensurable, so is FD also, and each of the straight lines AB, CD will be a fifth binomial. []
καὶ εἰ μὲν ἡ ΑΕ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ, καὶ ἡ ΓΖ σύμμετρός ἐστιν αὐτῇ, καί ἐστιν ἑκατέρα τετάρτη.
But, if neither of the straight lines AE, EB is so commensurable, neither of the straight lines CF, FD is commensurable with the rational straight line set out, and each of the straight lines AB, CD will be a sixth binomial. []
εἰ δὲ ἡ ΕΒ, καὶ ἡ ΖΔ, καὶ ἔσται ἑκατέρα πέμπτη. εἰ δὲ οὐδετέρα τῶν ΑΕ, ΕΒ, καὶ τῶν ΓΖ, ΖΔ οὐδετέρα σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ, καὶ ἔσται ἑκατέρα ἕκτη.
Hence a straight line commensurable in length with a binomial straight line is binomial and the same in order. Q. E. D.
ὥστε ἡ τῇ ἐκ δύο ὀνομάτων μήκει σύμμετρος ἐκ δύο ὀνομάτων ἐστὶ καὶ τῇ τάξει ἡ αὐτή· ὅπερ ἔδει δεῖξαι.
A straight line commensurable in length with a bimedial straight line is itself also bimedial and the same in order.
ἡ τῇ ἐκ δύο μέσων μήκει σύμμετρος καὶ αὐτὴ ἐκ δύο μέσων ἐστὶ καὶ τῇ τάξει ἡ αὐτή. ἔστω ἐκ δύο μέσων ἡ ΑΒ, καὶ τῇ ΑΒ σύμμετρος ἔστω μήκει ἡ ΓΔ·
Let AB be bimedial, and let CD be commensurable in length with AB; I say that CD is bimedial and the same in order with AB.
λέγω, ὅτι ἡ ΓΔ ἐκ δύο μέσων ἐστὶ καὶ τῇ τάξει ἡ αὐτὴ τῇ ΑΒ.
For, since AB is bimedial, let it be divided into its medials at E; therefore AE, EB are medial straight lines commensurable in square only. []
ἐπεὶ γὰρ ἐκ δύο μέσων ἐστὶν ἡ ΑΒ, διῃρήσθω εἰς τὰς μέσας κατὰ τὸ Ε· αἱ ΑΕ, ΕΒ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι.
And let it be contrived that, as AB is to CD, so is AE to CF; therefore also the remainder EB is to the remainder FD as AB is to CD. []
καὶ γεγονέτω ὡς ἡ ΑΒ πρὸς ΓΔ, ἡ ΑΕ πρὸς ΓΖ· καὶ λοιπὴ ἄρα ἡ ΕΒ πρὸς λοιπὴν τὴν ΖΔ ἐστιν, ὡς ἡ ΑΒ πρὸς ΓΔ. σύμμετρος δὲ ἡ ΑΒ τῇ ΓΔ μήκει·
But AB is commensurable in length with CD; therefore AE, EB are also commensurable with CF, FD respectively. []
σύμμετρος ἄρα καὶ ἑκατέρα τῶν ΑΕ, ΕΒ ἑκατέρᾳ τῶν ΓΖ, ΖΔ. μέσαι δὲ αἱ ΑΕ, ΕΒ·
But AE, EB are medial; therefore CF, FD are also medial. []
μέσαι ἄρα καὶ αἱ ΓΖ, ΖΔ.
And since, as AE is to EB, so is CF to FD, [] and AE, EB are commensurable in square only, CF, FD are also commensurable in square only. []
καὶ ἐπεί ἐστιν ὡς ἡ ΑΕ πρὸς ΕΒ, ἡ ΓΖ πρὸς ΖΔ, αἱ δὲ ΑΕ, ΕΒ δυνάμει μόνον σύμμετροί εἰσιν, καὶ αἱ ΓΖ, ΖΔ [ἄρα] δυνάμει μόνον σύμμετροί εἰσιν.
But they were also proved medial; therefore CD is bimedial.
ἐδείχθησαν δὲ καὶ μέσαι· ἡ ΓΔ ἄρα ἐκ δύο μέσων ἐστίν.
I say next that it is also the same in order with AB.
λέγω δή, ὅτι καὶ τῇ τάξει ἡ αὐτή ἐστι τῇ ΑΒ.
For since, as AE is to EB, so is CF to FD, therefore also, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle CF, FD; therefore, alternately, as the square on AE is to the square on CF, so is the rectangle AE, EB to the rectangle CF, FD. []
ἐπεὶ γάρ ἐστιν ὡς ἡ ΑΕ πρὸς ΕΒ, ἡ ΓΖ πρὸς ΖΔ, καὶ ὡς ἄρα τὸ ἀπὸ τῆς ΑΕ πρὸς τὸ ὑπὸ τῶν ΑΕΒ, οὕτως τὸ ἀπὸ τῆς ΓΖ πρὸς τὸ ὑπὸ τῶν ΓΖΔ· ἐναλλὰξ ὡς τὸ ἀπὸ τῆς ΑΕ πρὸς τὸ ἀπὸ τῆς ΓΖ, οὕτως τὸ ὑπὸ τῶν ΑΕΒ πρὸς τὸ ὑπὸ τῶν ΓΖΔ.
But the square on AE is commensurable with the square on CF; therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD.
σύμμετρον δὲ τὸ ἀπὸ τῆς ΑΕ τῷ ἀπὸ τῆς ΓΖ· σύμμετρον ἄρα καὶ τὸ ὑπὸ τῶν ΑΕΒ τῷ ὑπὸ τῶν ΓΖΔ.
If therefore the rectangle AE, EB is rational, the rectangle CF, FD is also rational, [and for this reason CD is a first bimedial]; [] but if medial, medial, [] and each of the straight lines AB, CD is a second bimedial. []
εἴτε οὖν ῥητόν ἐστι τὸ ὑπὸ τῶν ΑΕΒ, καὶ τὸ ὑπὸ τῶν ΓΖΔ ῥητόν ἐστιν [καὶ διὰ τοῦτό ἐστιν ἐκ δύο μέσων πρώτη]. εἴτε μέσον, μέσον, καί ἐστιν ἑκατέρα δευτέρα. καὶ διὰ τοῦτο ἔσται ἡ ΓΔ τῇ ΑΒ τῇ τάξει ἡ αὐτή·
And for this reason CD will be the same in order with AB. Q. E. D.
ὅπερ ἔδει δεῖξαι. ἡ τῇ μείζονι σύμμετρος καὶ αὐτὴ μείζων ἐστίν.
A straight line commensurable with a major straight line is itself also major.
ἔστω μείζων ἡ ΑΒ, καὶ τῇ ΑΒ σύμμετρος ἔστω ἡ ΓΔ·
Let AB be major, and let CD be commensurable with AB; I say that CD is major.
λέγω, ὅτι ἡ ΓΔ μείζων ἐστίν. διῃρήσθω ἡ ΑΒ κατὰ τὸ Ε·
Let AB be divided at E; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. [] Let the same construction be made as before.
αἱ ΑΕ, ΕΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων ῥητόν, τὸ δʼ ὑπʼ αὐτῶν μέσον· καὶ γεγονέτω τὰ αὐτὰ τοῖς πρότερον.
Then since, as AB is to CD, so is AE to CF, and EB to FD, therefore also, as AE is to CF, so is EB to FD. []
καὶ ἐπεί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἥ τε ΑΕ πρὸς τὴν ΓΖ καὶ ἡ ΕΒ πρὸς τὴν ΖΔ, καὶ ὡς ἄρα ἡ ΑΕ πρὸς τὴν ΓΖ, οὕτως ἡ ΕΒ πρὸς τὴν ΖΔ.
But AB is commensurable with CD; therefore AE, EB are also commensurable with CF, FD respectively. []
σύμμετρος δὲ ἡ ΑΒ τῇ ΓΔ. σύμμετρος ἄρα καὶ ἑκατέρα τῶν ΑΕ, ΕΒ ἑκατέρᾳ τῶν ΓΖ, ΖΔ.
And since, as AE is to CF, so is EB to FD, alternately also, as AE is to EB, so is CF to FD; [] therefore also, componendo, as AB is to BE, so is CD to DF; [] therefore also, as the square on AB is to the square on BE, so is the square on CD to the square on DF. []
καὶ ἐπεί ἐστιν ὡς ἡ ΑΕ πρὸς τὴν ΓΖ, οὕτως ἡ ΕΒ πρὸς τὴν ΖΔ, καὶ ἐναλλὰξ ὡς ἡ ΑΕ πρὸς ΕΒ, οὕτως ἡ ΓΖ πρὸς ΖΔ, καὶ συνθέντι ἄρα ἐστὶν ὡς ἡ ΑΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΓΔ πρὸς τὴν ΔΖ· καὶ ὡς ἄρα τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΕ, οὕτως τὸ ἀπὸ τῆς ΓΔ πρὸς τὸ ἀπὸ τῆς ΔΖ.
Similarly we can prove that, as the square on AB is to the square on AE, so also is the square on CD to the square on CF.
ὁμοίως δὴ δείξομεν, ὅτι καὶ ὡς τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΑΕ, οὕτως τὸ ἀπὸ τῆς ΓΔ πρὸς τὸ ἀπὸ τῆς ΓΖ.
Therefore also, as the square on AB is to the squares on AE, EB, so is the square on CD to the squares on CF, FD; therefore also, alternately, as the square on AB is to the square on CD, so are the squares on AE, EB to the squares on CF, FD. []
καὶ ὡς ἄρα τὸ ἀπὸ τῆς ΑΒ πρὸς τὰ ἀπὸ τῶν ΑΕ, ΕΒ, οὕτως τὸ ἀπὸ τῆς ΓΔ πρὸς τὰ ἀπὸ τῶν ΓΖ, ΖΔ· καὶ ἐναλλὰξ ἄρα ἐστὶν ὡς τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΓΔ, οὕτως τὰ ἀπὸ τῶν ΑΕ, ΕΒ πρὸς τὰ ἀπὸ τῶν ΓΖ, ΖΔ.
But the square on AB is commensurable with the square on CD; therefore the squares on AE, EB are also commensurable with the squares on CF, FD.
σύμμετρον δὲ τὸ ἀπὸ τῆς ΑΒ τῷ ἀπὸ τῆς ΓΔ· σύμμετρα ἄρα καὶ τὰ ἀπὸ τῶν ΑΕ, ΕΒ τοῖς ἀπὸ τῶν ΓΖ, ΖΔ.
And the squares on AE, EB together are rational; therefore the squares on CF, FD together are rational.
καί ἐστι τὰ ἀπὸ τῶν ΑΕ, ΕΒ ἅμα ῥητόν, καὶ τὰ ἀπὸ τῶν ΓΖ, ΖΔ ἅμα ῥητόν ἐστιν.
Similarly also twice the rectangle AE, EB is commensurable with twice the rectangle CF, FD.
ὁμοίως δὲ καὶ τὸ δὶς ὑπὸ τῶν ΑΕ, ΕΒ σύμμετρόν ἐστι τῷ δὶς ὑπὸ τῶν ΓΖ, ΖΔ.
And twice the rectangle AE, EB is medial; therefore twice the rectangle CF, FD is also medial. []
καί ἐστι μέσον τὸ δὶς ὑπὸ τῶν ΑΕ, ΕΒ· μέσον ἄρα καὶ τὸ δὶς ὑπὸ τῶν ΓΖ, ΖΔ.
Therefore CF, FD are straight lines incommensurable in square which make, at the same time, the sum of the squares on them rational, but the rectangle contained by them medial; therefore the whole CD is the irrational straight line called major. []
αἱ ΓΖ, ΖΔ ἄρα δυνάμει ἀσύμμετροί εἰσι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων ἅμα ῥητόν, τὸ δὲ δὶς ὑπʼ αὐτῶν μέσον· ὅλη ἄρα ἡ ΓΔ ἄλογός ἐστιν ἡ καλουμένη μείζων. ἡ ἄρα τῇ μείζονι σύμμετρος μείζων ἐστίν·
Therefore a straight line commensurable with the major straight line is major. Q. E. D.
ὅπερ ἔδει δεῖξαι.
A straight line commensurable with the side of a rational plus a medial area is itself also the side of a rational plus a medial area.
ἡ τῇ ῥητὸν καὶ μέσον δυναμένῃ σύμμετρος [καὶ αὐτὴ] ῥητὸν καὶ μέσον δυναμένη ἐστίν. ἔστω ῥητὸν καὶ μέσον δυναμένη ἡ ΑΒ, καὶ τῇ ΑΒ σύμμετρος ἔστω ἡ ΓΔ·
Let AB be the side of a rational plus a medial area, and let CD be commensurable with AB; it is to be proved that CD is also the side of a rational plus a medial area.
δεικτέον, ὅτι καὶ ἡ ΓΔ ῥητὸν καὶ μέσον δυναμένη ἐστίν. διῃρήσθω ἡ ΑΒ εἰς τὰς εὐθείας κατὰ τὸ Ε·
Let AB be divided into its straight lines at E; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. [] Let the same construction be made as before.
αἱ ΑΕ, ΕΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων μέσον, τὸ δʼ ὑπʼ αὐτῶν ῥητόν· καὶ τὰ αὐτὰ κατεσκευάσθω τοῖς πρότερον.
We can then prove similarly that CF, FD are incommensurable in square, and the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that the sum of the squares on CF, FD is also medial, and the rectangle CF, FD rational.
ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ΓΖ, ΖΔ δυνάμει εἰσὶν ἀσύμμετροι, καὶ σύμμετρον τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΕ, ΕΒ τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ, τὸ δὲ ὑπὸ ΑΕ, ΕΒ τῷ ὑπὸ ΓΖ, ΖΔ· ὥστε καὶ τὸ [μὲν] συγκείμενον ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ τετραγώνων ἐστὶ μέσον, τὸ δʼ ὑπὸ τῶν ΓΖ, ΖΔ ῥητόν. ῥητὸν ἄρα καὶ μέσον δυναμένη ἐστὶν ἡ ΓΔ·
Therefore CD is the side of a rational plus a medial area. Q. E. D.
ὅπερ ἔδει δεῖξαι. ἡ τῇ δύο μέσα δυναμένῃ σύμμετρος δύο μέσα δυναμένη ἐστίν.
A straight line commensurable with the side of the sum of two medial areas is the side of the sum of two medial areas.
ἔστω δύο μέσα δυναμένη ἡ ΑΒ, καὶ τῇ ΑΒ σύμμετρος ἡ ΓΔ·
Let AB be the side of the sum of two medial areas, and CD commensurable with AB; it is to be proved that CD is also the side of the sum of two medial areas.
δεικτέον, ὅτι καὶ ἡ ΓΔ δύο μέσα δυναμένη ἐστίν. ʼἐπεὶ γὰρ δύο μέσα δυναμένη ἐστὶν ἡ ΑΒ, διῃρήσθω εἰς τὰς εὐθείας κατὰ τὸ Ε·
For, since AB is the side of the sum of two medial areas, let it be divided into its straight lines at E; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and furthermore the sum of the squares on AE, EB incommensurable with the rectangle AE, EB. []
αἱ ΑΕ, ΕΒ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν [τετραγώνων] μέσον καὶ τὸ ὑπʼ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΕ, ΕΒ τετραγώνων τῷ ὑπὸ τῶν ΑΕ, ΕΒ· καὶ κατεσκευάσθω τὰ αὐτὰ τοῖς πρότερον.
Let the same construction be made as before.
ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ΓΖ, ΖΔ δυνάμει εἰσὶν ἀσύμμετροι καὶ σύμμετρον τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΕ, ΕΒ τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ, τὸ δὲ ὑπὸ τῶν ΑΕ, ΕΒ τῷ ὑπὸ τῶν ΓΖ, ΖΔ·
We can then prove similarly that CF, FD are also incommensurable in square, the sum of the squares on AE, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, EB with the rectangle CF, FD; so that the sum of the squares on CF, FD is also medial, the rectangle CF, FD is medial, and moreover the sum of the squares on CF, FD is incommensurable with the rectangle CF, FD. Therefore CD is the side of the sum of two medial areas. Q. E. D.
ὥστε καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ τετραγώνων μέσον ἐστὶ καὶ τὸ ὑπὸ τῶν ΓΖ, ΖΔ μέσον καὶ ἔτι ἀσύμμετρον τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΓΖ, ΖΔ τετραγώνων τῷ ὑπὸ τῶν ΓΖ, ΖΔ. ἡ ἄρα ΓΔ δύο μέσα δυναμένη ἐστίν· ὅπερ ἔδει δεῖξαι.
If a rational and a medial area be added together, four irrational straight lines arise, namely a binomial or a first bimedial or a major or a side of a rational plus a medial area.
ῥητοῦ καὶ μέσου συντιθεμένου τέσσαρες ἄλογοι γίγνονται ἤτοι ἐκ δύο ὀνομάτων ἢ ἐκ δύο μέσων πρώτη ἢ μείζων ἢ ῥητὸν καὶ μέσον δυναμένη. ἔστω ῥητὸν μὲν τὸ ΑΒ, μέσον δὲ τὸ ΓΔ·
Let. AB be rational, and CD medial; I say that thesideof the area AD is a binomial or a first bimedial or a major or a side of a rational plus a medial area.
λέγω, ὅτι ἡ τὸ ΑΔ χωρίον δυναμένη ἤτοι ἐκ δύο ὀνομάτων ἐστὶν ἢ ἐκ δύο μέσων πρώτη ἢ μείζων ἢ ῥητὸν καὶ μέσον δυναμένη. τὸ γὰρ ΑΒ τοῦ ΓΔ ἤτοι μεῖζόν ἐστιν ἢ ἔλασσον.