9.2. In triangle $ABC$, side $BC$ is equal to segment $AM$, where $M$ is the point of intersection of the medians. Find the angle $\angle BMC$.
Answer: $90^{\circ}$. Solution. Let $A N-$ be the median. By the property of the intersection point of medians, $M N=\frac{1}{2} A M=\frac{1}{2} B C$. Thus, in triangle $B M C$, the median $M N$ is equal to half of side $B C$. Therefore, $\triangle B M N$ is a right triangle with a right angle at $B M N$ (this is a known fact, which can be easily proven by considering isosceles triangles $B M N$ and $C M N$: the doubled sum of the angles at their base equals $180^{\circ}$ ).