Solution:

First, we check that $ABCD$ is cyclic. Reflect $B$ over $\overline{AC}$ to $B'$ on $\overline{AD}$, and note that $B'C = CD$. Therefore, $m \angle ADC = m \angle B'DC = m \angle CB'D = 180^{\circ} - m \angle AB'C = 180^{\circ} - m \angle CBA$.

Now $m \angle CBD = m \angle CAD = 20^{\circ}$ and $m \angle ADC = 180^{\circ} - m \angle CBA = 30^{\circ}$. Triangle $CDE$ is isosceles, so $m \angle CED = 75^{\circ}$ and $m \angle AEC = 105^{\circ}$. It follows that $m \angle ECA = 180^{\circ} - m \angle AEC - m \angle CAE = 55^{\circ}$.