8.3. In triangle $A B C$, the bisector $A M$ and the median $B N$ intersect at point $O$. It turns out that the areas of triangles $A B M$ and $M N C$ are equal. Find $\angle M O N$.
Answer: $90^{\circ}$. Solution. Since $S_{A M N}=S_{M N C}$ (because $A N=N C$), from the condition of the problem we have $S_{A B M}=S_{A N M}$. Therefore, in triangles $A B M$ and $A M N$, the heights drawn from vertices $B$ and $N$ are equal. Let $B_{1}$ and $N_{1}$ be the bases of these heights. Then the right triangles $A B B_{1}$ and $A N N_{1}$ are equal, since $B B_{1}=N N_{1}$ and $\angle B A B_{1}=\angle N A N_{1}$. Therefore, $A B=A N$ and thus points $B_{1}$ and $N_{1}$ coincide with each other and coincide with point $O$.