(related to Lemma: Lem. 10.028.2: Finding Two Squares With Sum Not Square)

- For let the (number created) from (multiplying) $AB$ and $BC$, as we said, be square.
- And (let) $CA$ (be) even.
- And let $CA$ have been cut in half at $D$.

- So it is clear that the square (number created) from (multiplying) $AB$ and $BC$, plus the square on $CD$, is equal to the square on $BD$ [see previous lemma].
- Let the unit $DE$ have been subtracted (from $BD$).
- Thus, the (number created) from (multiplying) $AB$ and $BC$, plus the (square) on $CE$, is less than the square on $BD$.
- I say, therefore, that the square (number created) from (multiplying) $AB$ and $BC$, plus the (square) on $CE$, is not 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 (than the square on $BE$), lest the unit be divided.
- First of all, if possible, let the (number created) from (multiplying) $AB$ and $BC$, plus the (square) on $CE$, be equal to the (square) on $BE$.
- And let $GA$ be double the unit $DE$.
- Therefore, since the whole of $AC$ is double the whole of $CD$, of which $AG$ is double $DE$, the remainder $GC$ is thus double the remainder $EC$.
- Thus, $GC$ has been cut in half at $E$.
- Thus, the (number created) from (multiplying) $GB$ and $BC$, plus the (square) on $CE$, is equal to the square on $BE$ [Prop. 2.6].
- But, the (number created) from (multiplying) $AB$ and $BC$, plus the (square) on $CE$, was also assumed (to be) equal to the square on $BE$.
- Thus, the (number created) from (multiplying) $GB$ and $BC$, plus the (square) on $CE$, is equal to the (number created) from (multiplying) $AB$ and $BC$, plus the (square) on $CE$.
- And subtracting the (square) on $CE$ from both, $AB$ is inferred (to be) equal to $GB$.
- The very thing is absurd.
- Thus, the (number created) from (multiplying) $AB$ and $BC$, plus the (square) on $CE$, is not equal to the (square) on $BE$.
- So I say that (it is) not less than the (square) on $BE$ either.
- For, if possible, let it be equal to the (square) on $BF$.
- And (let) $HA$ (be) double $DF$.
- And it can again be inferred that $HC$ (is) double $CF$.
- Hence, $CH$ has also been cut in half at $F$.
- And, on account of this, the (number created) from (multiplying) $HB$ and $BC$, plus the (square) on $FC$, becomes equal to the (square) on $BF$ [Prop. 2.6].
- And the (number created) from (multiplying) $AB$ and $BC$, plus the (square) on $CE$, was also assumed (to be) equal to the (square) on $BF$.
- Hence, the (number created) from (multiplying) $HB$ and $BC$, plus the (square) on $CF$, will also be equal to the (number created) from (multiplying) $AB$ and $BC$, plus the (square) on $CE$.
- The very thing is absurd.
- Thus, the (number created) from (multiplying) $AB$ and $BC$, plus the (square) on $CE$, is not equal to less than the (square) on $BE$.
- And it was shown that (is it) not equal to the (square) on $BE$ either.
- Thus, the (number created) from (multiplying) $AB$ and $BC$, plus the square on $CE$, is not square.
- (Which is) the very thing it was required to show.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"