It is then manifest that the square product of AB, BC together with the square on CD is equal to the square on BD. [See [Lemma 1](elem.10.28.l.1)]

Let the unit DE be subtracted; therefore the product of AB, BC together with the square on CE is less than the square on BD.

I say then that the square product of AB, BC together with the square on CE will not be square.

For, if it is square, it is either equal to the square on BE, or less than the square on BE, but cannot any more be greater, lest the unit be divided.

First, if possible, let the product of AB, BC together with the square on CE be equal to the square on BE, and let GA be double of the unit DE.

Since then the whole AC is double of the whole CD, and in them AG is double of DE, therefore the remainder GC is also double of the remainder EC; therefore GC is bisected by E.

Therefore the product of GB, BC together with the square on CE is equal to the square on BE. [[II. 6](elem.2.6)]

But the product of AB, BC together with the square on CE is also, by hypothesis, equal to the square on BE; therefore the product of GB, BC together with the square on CE is equal to the product of AB, BC together with the square on CE.

And, if the common square on CE be subtracted, it follows that AB is equal to GB: which is absurd.

Therefore the product of AB, BC together with the square on CE is not equal to the square on BE.

I say next that neither is it less than the square on BE.

For, if possible, let it be equal to the square on BF, and let HA be double of DF.

Now it will again follow that HC is double of CF; so that CH has also been bisected at F, and for this reason the product of HB, BC together with the square on FC is equal to the square on BF. [[II. 6](elem.2.6)]

But, by hypothesis, the product of AB, BC together with the square on CE is also equal to the square on BF.

Thus the product of HB, BC together with the square on CF will also be equal to the product of AB, BC together with the square on CE: which is absurd.

Therefore the product of AB, BC together with the square on CE is not less than the square on BE.

And it was proved that neither is it equal to the square on BE.

Therefore the product of AB, BC together with the square on CE is not square. Q. E. D.