1. Given a convex quadrilateral $A B C D$ in which $\angle B A C=20^{\circ}, \angle C A D=60^{\circ}$, $\angle A D B=50^{\circ}$, and $\angle B D C=10^{\circ}$. Find $\angle A C B$.
Solution. The triangle $\triangle A C D$ is equilateral since $\angle C A D=60^{\circ}$ and also $\angle C D A=$ $\angle C D B+\angle A D B=50^{\circ}+10^{\circ}=60^{\circ}$. Hence $A C=C D=A D$. Now the triangle $\triangle B A D$ is isosceles since $\angle A D B=\angle A B D=50^{\circ}$ and hence $A B=A D=A C$ from which we imply that $\triangle A B C$ is also isosceles. Thus
$$
\angle A B C=\angle A C B=\frac{180^{\circ}-20^{\circ}}{2}=80^{\circ} .
$$