Question 95
In a convex quadrilateral $ABCD$, the diagonals intersect at $E$, $BE = \sqrt{2} \cdot ED$ and $\angle BEC = 45^\circ$. Let $F$ be the foot of the perpendicular from $A$ to $BC$ and $P$ be the second intersection point of the circumcircle of triangle $BFD$ and the line segment $[DC]$. Find $\angle APD$.