Proof: By Euclid
(related to Proposition: Prop. 10.029: Construction of Rational Straight Lines Commensurable in Square When Square Differences Commensurable)
 For let some rational (straight line) $AB$ be laid down, and two square numbers, $CD$ and $DE$, such that the difference between them, $CE$, is not square [Prop. 10.28 lem. I] .
 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.
 Therefore, since as the (square) on $BA$ is to the (square) on $AF$, so $DC$ (is) to $CE$, the (square) on $BA$ thus has to the (square) on $AF$ the ratio which the number $DC$ (has) to the number $CE$.
 Thus, the (square) on $BA$ is commensurable with the (square) on on $AF$ [Prop. 10.6].
 And the (square) on $AB$ (is) rational [Def. 10.4] .
 Thus, the (square) on $AF$ (is) also rational.
 Thus, $AF$ (is) also rational.
 And since $DC$ does not have to $CE$ the ratio which (some) square number (has) to (some) square number, the (square) on $BA$ thus does not have to the (square) on $AF$ the ratio which (some) square number has to (some) square number either.
 Thus, $AB$ is incommensurable in length with $AF$ [Prop. 10.9].
 Thus, the rational (straight lines) $BA$ and $AF$ 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 $CD$ has to $DE$ the ratio which (some) square number (has) to (some) square number.
 Thus, the (square) on $AB$ also has to the (square) on $BF$ the ratio which (some) square number has to (some) square number.
 $AB$ is thus commensurable in length with $BF$ [Prop. 10.9].
 And the (square) on $AB$ is equal to the (sum of the squares) on $AF$ and $FB$ [Prop. 1.47].
 Thus, the square on $AB$ is greater than (the square on) $AF$ by (the square on) $BF$, (which is) commensurable (in length) with ($AB$).
 Thus, two rational (straight lines), $BA$ and $AF$, commensurable in square only, have been found such that the square on the greater, $AB$, is larger than (the square on) the lesser, $AF$, by the (square) on $BF$, (which is) commensurable in length with ($AB$).^{1}
 (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes