Proof: By Euclid
(related to Lemma: Lem. 13.13: Construction of Regular Tetrahedron within Given Sphere)
 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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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"