2. In $\triangle A B C$, $A B=A C, \angle A=20^{\circ}$, point $M$ is on $A C$ and satisfies $A M=B C$. Find the degree measure of $\angle B M C$.
(提示: 作正 $\triangle P A B$, 点 $P 、 M$ 位于 $A B$ 的异側.易证 $\triangle P A M \cong \triangle A C B, P M=A B=P A=P B$. 从而, $P$为 $\triangle A B M$ 的外心, $\angle A B M=\frac{1}{2} \angle A P M=\frac{1}{2} \times 20^{\circ}=$ $\left.10^{\circ}, \angle B M C=\angle A B M+\angle B A M=30^{\circ}.\right)$

(提示: Construct an equilateral $\triangle P A B$, with points $P$ and $M$ on opposite sides of $A B$. It is easy to prove that $\triangle P A M \cong \triangle A C B, P M=A B=P A=P B$. Therefore, $P$ is the circumcenter of $\triangle A B M$, $\angle A B M=\frac{1}{2} \angle A P M=\frac{1}{2} \times 20^{\circ}=$ $\left.10^{\circ}, \angle B M C=\angle A B M+\angle B A M=30^{\circ}.\right)$