Question 74
Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $E$ be the point on the side $BC$ such that $2 \angle BAE = \angle ACB$. Let $D$ be the second intersection of $AB$ and the circumcircle of the triangle $AEC$, and let $P$ be the second intersection of $CD$ and the circumcircle of the triangle $DBE$. Compute the angle $\angle BAP$.