Olympiad Geometry · Superchallenge problem

In $\triangle ABC$, $m\angle A = 40^\circ$ and $\overline{BC}$ is the shortest side. Point $D$ lies on $\overline{AB}$ with $BD = BC$, and point $E$ lies on $\overline{AC}$ with $CE = CB$. Segments $\overline{CD}$ and $\overline{BE}$ meet at $P$. Find $m\angle BPC$.