Prove that triangle $EDC$ is similar to triangle $EAB$, given $CD$ is parallel to $AB$

Question

Prove that triangle $EDC$ is similar to triangle $EAB$, given $CD$ is parallel to $AB$, and specify a reason for similarity.

I deduce that triangles $ACE$ and $CDE$ have common height, triangle $EAB$ and triangle $EDB$ have a common height, $ED/EA$ is known.

Answer 1

Apply alternate angle theorem to lines $AB, CD$ and $AE$. Thus, $\angle CDE \cong \angle BAD$. Similarly, $\angle DCE \cong \angle ABE$. This is enough to declare similarity based on that three angles in one triangle are congruent to three angles in another triangle. You can also use that vertical angles are congruent.