8.3. In triangle $A B C$, side $A C$ is the largest. Points $M$ and $N$ on side $A C$ are such that $A M=A B$ and $C N=C B$. It is known that angle $N B M$ is three times smaller than angle $A B C$. Find $\angle A B C$.
Answer: $108^{\circ}$. Solution: Let $\angle N B M=x$. Then $\angle A B M+\angle N B C=\angle A B C+\angle N B M=4 x . \quad$ On the other hand, $\angle A B M+\angle N B C=\angle B M N+\angle B N M=180^{\circ}-\angle N B M=180^{\circ}-x$. Therefore, we have the equation $4 x=180^{\circ}-x$. Thus, $5 x=180^{\circ}, x=36^{\circ}$ and $\angle A B C=3 x=108^{\circ}$.