Kitap 10
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.
54
And, since AE is incommensurable in length with ED, while AE is commensurable with AG, and DE is commensurable with EF, therefore AG is also incommensurable with EF, [X. 13] so that AH is also incommensurable with EL. [VI. 1, X. 11]
54
But AH is equal to SN, and EL to MR; therefore SN is also incommensurable with MR.
