Proof: By Euclid

fig23e

fig23ae

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Let the straight lines $AB$ and $LO$ be set out, and let $AB$ be greater, and let the semicircle $ABC$ have been drawn around it.
And let $AC$, equal to the straight line $LO$, which is not greater than the diameter $AB$, have been inserted into the semicircle $ABC$ [Prop. 4.1].
And let $CB$ have been joined.
Therefore, since the angle $ACB$ is in the semicircle $ACB$, $ACB$ is thus a right angle [Prop. 3.31].
Thus, the (square) on $AB$ is equal to the (sum of the) squares on $AC$ and $CB$ [Prop. 1.47].
Hence, the (square) on $AB$ is greater than the (square) on $AC$ by the (square) on $CB$.
And $AC$ (is) equal to $LO$.
Thus, the (square) on $AB$ is greater than the (square) on $LO$ by the (square) on $CB$.
Therefore, if we take $OR$ equal to $BC$ then the (square) on $AB$ will be greater than the (square) on $LO$ by the (square) on $OR$.
(Which is) the very thing it was prescribed to do.
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