Proof: By Euclid

fig030e

Let the rational (straight line) $AB$ be laid out, and the two square numbers, $CE$ and $ED$, such that the sum of them, $CD$, is not square [Prop. 10.28 lem. II] .
And let the semicircle $AFB$ have been drawn on $AB$.
And let it be contrived that as $DC$ (is) to $CE$, so the (square) on $BA$ (is) to the (square) on $AF$ [Prop. 10.6 corr.] .
And let $FB$ have been joined.
So, similarly to the (proposition) before this, we can show that $BA$ and $AF$ are rational (straight lines which are) commensurable in square only.
And since as $DC$ is to $CE$, so the (square) on $BA$ (is) to the (square) on $AF$, thus, via convertion, as $CD$ (is) to $DE$, so the (square) on $AB$ (is) to the (square) on $BF$ [Prop. 5.19 corr.] 2, [Prop. 3.31], [Prop. 1.47].
And $CD$ does not have to $DE$ the ratio which (some) square number (has) to (some) square number.
Thus, the (square) on $AB$ does not have to the (square) on $BF$ the ratio which (some) square number has to (some) square number either.
Thus, $AB$ is incommensurable in length with $BF$ [Prop. 10.9].
And the square on $AB$ is greater than the (square on) $AF$ by the (square) on $FB$ [Prop. 1.47], (which is) incommensurable (in length) with ($AB$).
Thus, $AB$ and $AF$ are rational (straight lines which are) commensurable in square only, and the square on $AB$ is greater than (the square on) $AF$ by the (square) on $FB$, (which is) incommensurable (in length) with $(AB)$.¹
(Which is) the very thing it was required to show.

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$AB$ and $AF$ have lengths $1$ and $1/\sqrt{1+k^2}$ times that of $AB$, respectively, where $k=\sqrt{DE/CE}$ (translator's note) ↩