jmc

Angles $\angle DCB$ and $\angle B$ are alternate interior angles, so they are congruent. Therefore, $m\angle B=40^{\circ }$.

Since $AC=BC$, triangle $△ABC$ is isosceles with equal angles at $A$ and $B$. Therefore, $m\angle A=40^{\circ }$.

Finally, $\angle A$ and $\angle ECD$ are corresponding angles, so $m\angle ECD=m\angle A=\boxed{40}$ degrees.