$\angle BAC = \pi/2$, $A', B', C'$ are on the sides or on the extensions of the sides $BC, CA, AB$ respectively, $\triangle ABC \sim \triangle A'B'C'$.
Prove (or disprove) that $A'$ is the midpoint of $BC$.
For a hint, see the comment under this question.
It is true!
$C'A'B'A$ can be inscribed in a circle, hence $\angle A'AB'=\angle A'C'B'=\angle A'CB'$. This implies that $AA'=A'C$
Similarly, you get that $AA'=BA'$ by looking at another pair of equal angles. QED