Right-angled Triangle and Triangles Inside it

Question

Let $ABC$ be a right-angled triangle with |AB|=|BC|. Let $D$ and $E$ be points on the side $BC$ satisfying $|BD|=|CE|$. Let $P$ be the intersection point of the line through $B$ perpendicular to $AD$ with the side $AC$. Prove that $\angle PEC=\angle ADB$

Any hints?

Thanks

Answer 1

Solution: Flip triangle $ABC$ across $AC$ to obtain the square $ABCB'$ (i.e. point $B'$ is the symmetric image of $B$ with respect to reflection in line $AC$). Extend $BP$ until it intersects the edge $CB'$ of the square $ABCB'$ (edge $CB'$ is parallel and equal to edge $AB$) and denote the point of intersection $E' = BP \cap CB'$.

1. Triangles $ABD$ and $BCE'$ are congruent because $AB = BC, \,\, \angle \, ABD = \angle \, BCE' = 90^{\circ}$ and $\angle \, BAD = \angle \, CBE'$ due to the fact that $AB$ is perpendicular to $BC$ and $AD$ is perpendicular to $BE'$.

2. Consequently, $\angle\, BDA = \angle \, CE'B = \angle \, CE'P = \alpha$ and $CE' = BD = CE$.

3. Furthermore, triangles $EPC$ and $E'PC$ are congruent because $CE = CE'$, $\,\, \angle \, ECP = \angle E'CP = 45^{\circ}$ and $CP$ is a common segment.

4. Therefore, $\angle \, CEP = \angle \, CE'P = \alpha = \angle \, ADB$.