Angling-Triangling!

Given $\triangle ABC$ , let $D$ be a point on $AB$ produced beyond $B$ , such that $BD=BC$ , and let $E$ be a point on $AC$ produced beyond $C$ , such that $CE=BC$ . Let $P$ be the intersection of $BE$ and $CD$ , and suppose that:

$\large{\dfrac{DP}{BE}+ \dfrac{EP}{CD} = 2\sin \left( \dfrac{\angle BAC}{2} \right) }$

Submit the value of $\angle BAC$ in degrees as your answer.