Answer: $\mathbf{5 5}$. First, we check that $A B C D$ is cyclic. Reflect $B$ over $\overline{A C}$ to $B^{\prime}$ on $\overline{A D}$, and note that $B^{\prime} C=C D$. Therefore, $m \angle A D C=m \angle B^{\prime} D C=m \angle C B^{\prime} D=180^{\circ}-m \angle A B^{\prime} C=180^{\circ}-m \angle C B A$. Now $m \angle C B D=m \angle C A D=20^{\circ}$ and $m \angle A D C=180^{\circ}-m \angle C B A=30^{\circ}$. Triangle $C D E$ is isosceles, so $m \angle C E D=75^{\circ}$ and $m \angle A E C=105^{\circ}$. It follows that $m \angle E C A=180^{\circ}-m \angle A E C-m \angle C A E=55^{\circ}$.
8. [6] $A B C D$ is a convex quadrilateral such that $A B<\underline{A D}$. The diagonal $\overline{A C}$ bisects $\angle B A D$, and $m \angle A B D=130^{\circ}$. Let $E$ be a point on the interior of $\overline{A D}$, and $m \angle B A D=40^{\circ}$. Given that $B C=$ $C D=D E$, determine $m \angle A C E$ in degrees.