Proof: By Euclid
(related to Proposition: 6.08: Perpendicular in RightAngled Triangle makes two Similar Triangles)
 For since (angle) $BAC$ is equal to $ADB$  for each (are) right angles  and the (angle) at $B$ (is) common to the two triangles $ABC$ and $ABD$, the remaining (angle) $ACB$ is thus equal to the remaining (angle) $BAD$ [Prop. 1.32].
 Thus, triangle $ABC$ is equiangular to triangle $ABD$.
 Thus, as $BC$, subtending the right angle in triangle $ABC$, is to $BA$, subtending the right angle in triangle $ABD$, so the same $AB$, subtending the angle at $C$ in triangle $ABC$, (is) to $BD$, subtending the equal (angle) $BAD$ in triangle $ABD$, and, further, (so is) $AC$ to $AD$, (both) subtending the angle at $B$ common to the two triangles [Prop. 6.4].
 Thus, triangle $ABC$ is equiangular to triangle $ABD$, and has the sides about the equal angles proportional.
 Thus, triangle $ABC$ [is] [similar]bookofproofs$1983 to triangle $ABD$ [Def. 6.1] .
 So, similarly, we can show that triangle $ABC$ is also similar to triangle $ADC$.
 Thus, [triangles] $ABD$ and $ADC$ are each similar to the whole (triangle) $ABC$.
 So I say that triangles $ABD$ and $ADC$ are also similar to one another.
 For since the right angle $BDA$ is equal to the right angle $ADC$, and, indeed, (angle) $BAD$ was also shown (to be) equal to the (angle) at $C$, thus the remaining (angle) at $B$ is also equal to the remaining (angle) $DAC$ [Prop. 1.32].
 Thus, triangle $ABD$ is equiangular to triangle $ADC$.
 Thus, as $BD$, subtending (angle) $BAD$ in triangle $ABD$, is to $DA$, subtending the (angle) at $C$ in triangle $ADC$, (which is) equal to (angle) $BAD$, so (is) the same $AD$, subtending the angle at $B$ in triangle $ABD$, to $DC$, subtending (angle) $DAC$ in triangle $ADC$, (which is) equal to the (angle) at $B$, and, further, (so is) $BA$ to $AC$, (each) subtending right angles [Prop. 6.4].
 Thus, triangle $ABD$ is similar to triangle $ADC$ [Def. 6.1] .
 Thus, if, in a rightangled triangle, a (straight line) is drawn from the right angle perpendicular to the base then the triangles around the perpendicular are similar to the whole (triangle), and to one another.
 (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"