8.2. In triangle $A B C$, angle $A$ is the largest. Points $M$ and $N$ are symmetric to vertex $A$ with respect to the angle bisectors of angles $B$ and $C$ respectively. Find $\angle A$, if $\angle M A N=50^{\circ}$.
Answer: $80^{\circ}$. Solution. Let $\angle A=\alpha, \angle B=\beta, \angle C=\gamma$. Then $\angle A M B=90^{\circ}-\frac{\beta}{2}, \angle A N C=90^{\circ}-\frac{\gamma}{2}$ (since triangles $A M B$ and $\quad$ are isosceles). Therefore, $\angle M A N=180^{\circ}-\left(90^{\circ}-\frac{\beta}{2}\right)-\left(90^{\circ}-\frac{\gamma}{2}\right)=\frac{\beta+\gamma}{2}=90^{\circ}-\frac{\alpha}{2}$. From the problem statement, $90^{\circ}-\frac{\alpha}{2}=50^{\circ}$, hence, $\alpha=80^{\circ}$.