For, since AD is a first binomial straight line, let it be divided into its terms at E, and let AE be the greater term.
It is then manifest that 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 commensurable with AE,
and AE is commensurable in length with the rational straight line AB set out. [X. Deff. II. 1]
Let ED be bisected at the point F.
Then, since the square on AE is greater than the square on ED by the square on a straight line commensurable with AE,
therefore, if there be applied to the greater AE a parallelogram equal to the fourth part of the square on the less,
that is, to the square on EF, and deficient by a square figure, it divides it into commensurable parts. [X. 17]
Let then the rectangle AG, GE equal to the square on EF be applied to AE;
therefore AG is commensurable in length with EG.
Let GH, EK, FL be drawn from G, E, F parallel to either of the straight lines AB, CD;
let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, [II. 14]
and let them be placed so that MN is in a straight line with NO;
therefore RN is also in a straight line with NP.
And let the parallelogram SQ be completed;
therefore SQ is a square. [Lemma]
Now, since the rectangle AG, GE is equal to the square on EF, therefore, as AG is to EF, so is FE to EG; [VI. 17]
therefore also, as AH is to EL, so is EL to KG; [VI. 1]
therefore EL is a mean proportional between AH, GK.
But AH is equal to SN, and GK to NQ;
therefore EL is a mean proportional between SN, NQ.
But MR is also a mean proportional between the same SN, NQ;
[Lemma] therefore EL is equal to MR, so that it is also equal to PO.
But AH, GK are also equal to SN, NQ;
therefore the whole AC is equal to the whole SQ, that is, to the square on MO;
therefore MO is the “side” of AC.
I say next that MO is binomial.