Two altitudes CE and BD are drawn in a triangle ABC. Prove that angle AED= angle BCA

Question

I managed to prove that $\measuredangle ABD = \measuredangle ECD$. I know that if I prove that $\measuredangle EDB= \measuredangle ECB$, I can do the problem, but I don't know how.

Answer 1

Since $$\angle BEC = \angle BDC = 90^{\circ}$$ we see that $BCDE$ is cyclic.

So $$180^{\circ} -\angle AED= \angle DEB = 180^{\circ} -\angle DCB$$