Book 10
54
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]
54
Let then the rectangle AG, GE equal to the square on EF be applied to AE; therefore AG is commensurable in length with EG.
54
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.
54
And let the parallelogram SQ be completed; therefore SQ is a square. [Lemma]
54
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.
54
But AH is equal to SN, and GK to NQ; therefore EL is a mean proportional between SN, NQ.
54
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.
54
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.
54
I say next that MO is binomial.
54
For, since AG is commensurable with GE, therefore AE is also commensurable with each of the straight lines AG, GE. [X. 15]
54
But AE is also, by hypothesis, commensurable with AB; therefore AG, GE are also commensurable with AB. [X. 12]
54
And AB is rational; therefore each of the straight lines AG, GE is also rational; therefore each of the rectangles AH, GK is rational, [X. 19] and AH is commensurable with GK.
54
But AH is equal to SN, and GK to NQ; therefore SN, NQ, that is, the squares on MN, NO, are rational and commensurable.
