How to prove a triangle similarity problem

Question

If I have a triangle $ABC$ with point $E$ lying on $BC$ and point $D$ lying on $AB$ where $AE$ is the height to $BC$ and $CD$ is the height to $AB$, how can I prove that triangle $ABC$ is similar to triangle $BED$?

Answer 1

Since $CD \perp AB$ and $AE \perp BC$, we have that $ADEC$ to be a cyclic quadrilateral. This means $\angle{DCE} = \angle{DAE}$ and $\angle{DCA} = \angle{DEA}$. This means $$\angle{C} = \angle{DCE} + \angle{DCA} = \underbrace{\angle{DAE} + \angle{DEA} = \angle{EDB}}_{\text{Exterior angle of }\Delta ADE}$$ By a similar argument, $\angle{A} = \angle{BED}$.

Further $\angle{B}$ is common to $\Delta ABC$ and $\Delta EBD$.

Hence, the two triangles are similar.