Proof: By Euclid
(related to Proposition: Prop. 10.034: Construction of Components of Side of Rational plus Medial Area)
 Let the two medial (straight lines) $AB$ and $BC$, (which are) commensurable in square only, be laid out having the (rectangle contained) by them rational, (and) such that the square on $AB$ is greater than (the square on) $BC$ by the (square) on (some straight line) incommensurable (in length) with ($AB$) [Prop. 10.31].
 And let the semicircle $ADB$ have been drawn on $AB$.
 And let $BC$ have been cut in half at $E$.
 And let a (rectangular) parallelogram equal to the (square) on $BE$, (and) falling short by a square figure, have been applied to $AB$, (and let it be) the (rectangle contained by) $AFB$ [Prop. 6.28].
 Thus, $AF$ [is] [incommensurable in length]bookofproofs$1095 with $FB$ [Prop. 10.18].
 And let $FD$ have been drawn from $F$ at right angles to $AB$.
 And let $AD$ and $DB$ have been joined.
 Since $AF$ is incommensurable (in length) with $FB$, the (rectangle contained) by $BA$ and $AF$ is thus also incommensurable with the (rectangle contained) by $AB$ and $BF$ [Prop. 10.11].
 And the (rectangle contained) by $BA$ and $AF$ (is) equal to the (square) on $AD$, and the (rectangle contained) by $AB$ and $BF$ to the (square) on $DB$ [Prop. 10.32 lem.] .
 Thus, the (square) on $AD$ is also incommensurable with the (square) on on $DB$.
 And since the (square) on $AB$ is medial, the sum of the (squares) on $AD$ and $DB$ (is) thus also medial [Prop. 3.31], [Prop. 1.47].
 And since $BC$ is double $DF$ [see previous proposition], the (rectangle contained) by $AB$ and $BC$ (is) thus also double the (rectangle contained) by $AB$ and $FD$.
 And the (rectangle contained) by $AB$ and $BC$ (is) rational.
 Thus, the (rectangle contained) by $AB$ and $FD$ (is) also rational [Prop. 10.6], [Def. 10.4] .
 And the (rectangle contained) by $AB$ and $FD$ (is) equal to the (rectangle contained) by $AD$ and $DB$ [Prop. 10.32 lem.] .
 And hence the (rectangle contained) by $AD$ and $DB$ is rational.
 Thus, two straight lines, $AD$ and $DB$, (which are) incommensurable in square, have been found, making the sum of the squares on them medial, and the (rectangle contained) by them rational.^{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