Proof: By Euclid

Let the perpendicular $AD$ have been drawn [Prop. 1.12].
Therefore, since, in the right-angled triangle $ABC$, the (straight line) $AD$ has been drawn from the right angle at $A$ perpendicular to the base $BC$, the triangles $ABD$ and $ADC$ about the perpendicular are similar to the whole (triangle) $ABC$, and to one another [Prop. 6.8].
And since $ABC$ is similar to $ABD$, thus as $CB$ is to $BA$, so $AB$ (is) to $BD$ [Def. 6.1] .
And since three straight lines are proportional, as the first is to the third, so the figure (drawn) on the first is to the similar, and similarly described, (figure) on the second [Prop. 6.19 corr.] 4.
Thus, as $CB$ (is) to $BD$, so the figure (drawn) on $CB$ (is) to the similar, and similarly described, (figure) on $BA$.
And so, for the same (reasons), as $BC$ (is) to $CD$, so the figure (drawn) on $BC$ (is) to the (figure) on $CA$.
Hence, also, as $BC$ (is) to $BD$ and $DC$, so the figure (drawn) on $BC$ (is) to the (sum of the) similar, and similarly described, (figures) on $BA$ and $AC$ [Prop. 5.24].
And $BC$ is equal to $BD$ and $DC$.
Thus, the figure (drawn) on $BC$ (is) also equal to the (sum of the) similar, and similarly described, figures on $BA$ and $AC$ [Prop. 5.9].
Thus, in right-angled triangles, the figure (drawn) on the side subtending the right angle is equal to the (sum of the) similar, and similarly described, figures on the sides surrounding the right angle.
(Which is) the very thing it was required to show.
∎

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