In acute triangle $A B C$, the three altitudes meet at $H$. Given that $A H=B C$, calculate at least one of the angles of $\triangle A B C$.

Answer: $\angle A=45^{\circ}$ (angles $B$ and $C$ cannot be determined).
In the diagram, $\angle A F H=\angle C F B$ (both are right angles) and $\angle F A H=\angle F C B$ (both are complementary to $\angle A B C$ ). We get $\triangle A F H \cong \triangle C F B$ by AAS. It follows that $A F=F C$, so $\triangle A F C$ is right isosceles. We find that $\angle B A C=45^{\circ}$.

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