Proof: By Euclid

fig13e

For, let the figure of the semicircle have been set out, and let $DB$ have been joined.
And let the square $EC$ have been described on $AC$.
And let the parallelogram $FB$ have been completed.
Therefore, since, on account of triangle $DAB$ being equiangular to triangle $DAC$ [Prop. 6.8], [Prop. 6.4], (proportionally) as $BA$ is to $AD$, so $DA$ (is) to $AC$, the (rectangle contained) by $BA$ and $AC$ is thus equal to the (square) on $AD$ [Prop. 6.17].
And since as $AB$ is to $BC$, so $EB$ (is) to $BF$ [Prop. 6.1].
And $EB$ is the (rectangle contained) by $BA$ and $AC$ - for $EA$ (is) equal to $AC$.
And $BF$ the (rectangle contained) by $AC$ and $CB$.
Thus, as $AB$ (is) to $BC$, so the (rectangle contained) by $BA$ and $AC$ (is) to the (rectangle contained) by $AC$ and $CB$.
And the (rectangle contained) by $BA$ and $AC$ is equal to the (square) on $AD$, and the (rectangle contained) by $ACB$ (is) equal to the (square) on $DC$.
For the perpendicular $DC$ is in mean proportion to the pieces of the base, $AC$ and $CB$, on account of $ADB$ being a right angle [Prop. 6.8 corr.] .
Thus, as $AB$ (is) to $BC$, so the (square) on $AD$ (is) to the (square) on $DC$.
(Which is) the very thing it was required to show.
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