<s>[INST] ### Problem
Let $\triangle ABC$ be an acute triangle with $\angle BAC = 90^{\circ}$. Let $H$ be the foot of the altitude from $A$ to $BC$ and $M$ be the midpoint of $BC$. Prove that $\angle BHM = 90^{\circ}$.

### Proof
We have that $BM = MC$. Since $\triangle AHB \sim \triangle ABC$, we have $$\frac{BH}{AB} = \frac{AH}{AC}.$$But $AH = BM$, so $$\frac{BH}{AB} = \frac{BM}{AC}.$$So, $$\frac{AB}{AC} = \frac{BM}{BH}.$$Thus, $$\frac{AB^2}{AC^2} = \frac{BM^2}{BH^2}.$$We have $AB^2 + BC^2 = AC^2$ and $BH^2 + HM^2 = BM^2$. Thus, $$\frac{AC^2 - BC^2}{AC^2} = \frac{BM^2 - BH^2}{BM^2}.$$Since $2BM = BC$, we have $$\frac{AC^2 - BC^2}{AC^2} = \frac{BC^2 - 4BH^2}{BC^2}.$$Cross-multiplying and simplifying, we get $$AC^2 \cdot BC^2 - 4AC^2 \cdot BH^2 = AC^2 \cdot BC^2 - BC^4.$$Simplifying further, we get $$BC^4 = 4AC^2 \cdot BH^2.$$Since $BC^2 = 4BM^2$, we have $$16BM^2 \cdot BH^2 = 4AC^2 \cdot BH^2.$$Dividing both sides by $BH^2$, we get $$4BM^2 = AC^2.$$We have $AC^2 = AB^2 + BC^2$. Thus, $$4BM^2 = AB^2 + BC^2.$$We have $BM = \frac{1}{2}BC$. Thus, $$BC^2 = AB^2 + BC^2.$$Subtracting $BC^2$ from both sides, we get $$0 = AB^2.$$So, $AB = 0$. Thus, $B$ coincides with $A$, which is impossible. Therefore, our assumption that $\angle BHM$ is not a right angle is false. Thus, $\angle BHM = 90^{\circ}$.

### Alternative solutions are always welcome. If you have a different, elegant solution to this problem, please <a class="external text" href="https://artofproblemsolving.com/wiki/index.php?[/INST]