Example 1. As shown in the figure, in $\triangle A B C$, $A C=B C, \angle C$ $=20^{\circ}$. Also, $M$ is on $A C$, $N$ is on $B C$, and it satisfies $\angle B A N=50^{\circ}, \angle A B M=60^{\circ}$. Then $\angle N M B=$ $\qquad$ . (1991, Qinfen Cup Mathematics Invitational Competition)
Draw a line through $M$ parallel to $AB$ intersecting $BC$ at $K$. Connect $AK$ intersecting $MB$ at $P$, and connect $PN$. It is easy to know that $\triangle A P R$ and $\triangle M K P$ are both equilateral triangles.
$$
\begin{array}{l}
\because \angle B A N=50^{\circ}, \quad A C=B C, \angle C=20^{\circ}, \\
\therefore \angle A N B=50^{\circ}, \therefore B N=A B=B P . \\
\therefore \angle B P N=\angle B N P=80^{\circ}, \\
\therefore \angle K P N=40^{\circ}, \\
\angle P K N=180^{\circ}-60^{\circ}-80^{\circ}=40^{\circ} .
\end{array}
$$

Thus $P N=K N$.
$$
\therefore \triangle M P N \cong \triangle M K N \text {. }
$$

Therefore, $\angle P M N=\angle K M N$.
$$
\therefore \angle N M B=\frac{1}{2} \angle K M B=30^{\circ} \text {. }
$$