4. In quadrilateral $A B C D, \angle D A C=98^{\circ}, \angle D B C=82^{\circ}, \angle B C D=70^{\circ}$, and $B C=A D$. Find $\angle A C D$.
Answer: 28 Let $B^{\prime}$ be the reflection of $B$ across $C D$. Note that $A D=B C$, and $\angle D A C+\angle C B^{\prime} D=$ $180^{\circ}$, so $A C B^{\prime} D$ is a cyclic trapezoid. Thus, $A C B^{\prime} D$ is an isosceles trapezoid, so $\angle A C B^{\prime}=98^{\circ}$. Note that $\angle D C B^{\prime}=\angle B C D=70^{\circ}$, so $\angle A C D=\angle A C B^{\prime}-\angle D C B^{\prime}=98^{\circ}-70^{\circ}=28^{\circ}$.