Show that the following two angles are equal:

Question

$CE$ and $BD$ are two heights of triangle $ABC$. Show that the following two angles are equal: $\angle EDB=\angle ECB$

It's obvious that $\angle EBD=\angle DCE$ and also $\angle DCE + \angle ECB + \angle DBC =90^\circ$ but how to proceed??!!

Answer 1

Since $\angle{BEC}=\angle{BDC}=90^\circ$, the four points $B,C,D$ and $E$ are on the circle whose diameter is the side $BC$. (see, for example, here) Now, $\angle{EDB}=\angle{ECB}$ follows.