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Segments $A M$ and $B H$ are the median and altitude of an acute triangle $A B C$, respectively. It is known that $A H=1$ and $2 \angle M A C=\angle M C A$. Find the side $B C$.
Let $\angle M A C=\alpha$. Then $\angle M C A=2 \alpha$. Segment $H M$ is the median of the right triangle $B H C$, drawn from the vertex of the right angle. Therefore, $H M=M C$. Triangle $C M H$ is isosceles, so $\angle C H M=\angle M C H=2 \alpha$.

By the external angle theorem of a triangle, $\angle A H M=\angle C H M-\angle M A C=\alpha$.

Therefore, triangle $A H M$ is also isosceles, $H M=A H=1$. Consequently, $B C=2 H M=2$.

## Answer

2.