9.2. In triangle $A B C$, the median $B M$ is twice as small as the bisector $A N$. It is known that angle $C B M$ is three times larger than angle $C A N$. Find the angles of triangle $A B C$.
# Answer: $\angle A=\angle C=36^{\circ}, \angle B=108^{\circ}$. Solution.

Let $\alpha$ be the angle between the angle bisector $A N$ and the sides $A B$ and $A C$. Draw a line through point $M$ parallel to the angle bisector $A N$, and let $K$ be the point of intersection of this line with side $B C$. Since $M$ is the midpoint of $A C$, in triangle $A N C$, segment $M K$ is the midline.

![](https://cdn.mathpix.com/cropped/2024_05_06_92125968fdbb1ee934d2g-2.jpg?height=360&width=826&top_left_y=1079&top_left_x=1049)
Therefore, $M K=\frac{1}{2} A N=B M$ (by the problem's condition). Thus, triangle $B M K$ is isosceles, and hence, $\angle M K B=\angle M B K=3 \alpha$ (as the base angles). The angle $M K B$ is the exterior angle for triangle $M K C$, so $\angle M K B=\angle K M C+\angle K C M$, i.e., $3 \alpha=\alpha+\angle K C M$. Therefore, $\angle K C M=2 \alpha=\angle B A C$. Thus, we find that triangle $A B C$ is isosceles, and then the median $B M$ is also the altitude, i.e., triangle $B M C$ is a right triangle, and the sum of its acute angles $3 \alpha+2 \alpha=90^{\circ} \Leftrightarrow \alpha=18^{\circ}$. As a result, we get: $\angle A=\angle C=2 \alpha=36^{\circ}, \angle B=2 \cdot 3 \alpha=108^{\circ}$.