8.3. Given a triangle $A B C$. On the side $A C$, the largest in the triangle, points $M$ and $N$ are marked such that $A M=A B$ and $C N=C B$. It turns out that angle $N B M$ is three times smaller than angle $A B C$. Find $\angle A B C$.
Answer: $108^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_cad01b0e5b8f6cb6848bg-1.jpg?height=318&width=607&top_left_y=1780&top_left_x=245)

Solution. Point $N$ lies between $A$ and $M$, since $A M+C V=A B+B C > AC$. Let $\angle N B M = x$, then $\angle A B M + \angle N B C = \angle A B C + \angle N B M = 3x + x = 4x$. According to the problem, triangles $A B M$ and $N B C$ are isosceles, and therefore $\angle A B M + \angle N B C = \angle B M N + \angle B N M = 180^{\circ} - \angle N B M = 180^{\circ} - x$. Thus, we have the equation $4x = 180^{\circ} - x$, from which $x = 36^{\circ}$ and $\angle A B C = 3x = 108^{\circ}$.