9.2. The bisectors of the external angles $B$ and $C$ of triangle $ABC$ intersect at point $M$. a) Can angle $BMC$ be obtuse? b) Find angle $BAC$ if it is known that $\angle BMC = \frac{\angle BAM}{2}$.
Answer: a) cannot; b) $120^{\circ}$. Hint. Let $\angle A=\alpha, \angle B=\beta, \angle C=\gamma$. Then $\angle B M C=180^{\circ}-\left(\frac{180^{\circ}-\beta}{2}+\frac{180^{\circ}-\gamma}{2}\right)=\frac{\beta+\gamma}{2}=90^{\circ}-\frac{\alpha}{2}$. a) Therefore, angle $B M C$ cannot be obtuse. b) from the equation $\frac{\alpha}{2}=90^{\circ}-\frac{\alpha}{4}$ we get the value $\alpha=120^{\circ}$ (here we used the fact that point $M$ must lie on the bisector of angle $A$ ).