1. $\angle Q M C+\angle P M B=180^{\circ}-\angle C M B=80^{\circ}$. From the isosceles triangles $C Q M$ and $B P M$, we get that $\angle Q C M=$ $\angle Q M C$ and $\angle P B M=\angle P M B$. Therefore, $\angle Q C M+\angle P B M=80^{\circ}$. From triangle $C M B$, we also have that $\angle M C B+$ $\angle M B C=180^{\circ}-\angle C M B=80^{\circ}$. Thus, $\angle A C B+\angle A B C=80^{\circ}+80^{\circ}=160^{\circ}$ and $\angle C A B=180^{\circ}-\angle A C B-\angle A B C=20^{\circ}$.
1. Inside triangle $A B C$, a point $M$ is taken such that $\angle C M B=100^{\circ}$. The perpendicular bisectors of $B M$ and $C M$ intersect sides $A B$ and $A C$ at points $P$ and $Q$, respectively. Points $P, Q$, and $M$ lie on the same line. Find the measure of $\angle C A B$.