Let $ABCD$ be the trapezium with bases $AB$ and $CD$, such that $\sphericalangle ABC = 90^{\circ}$. The bisector of angle $BAD$ intersects the segment $BC$ in the point $P$. Show that if $\sphericalangle APD = 45^{\circ}$, then area of quadrilateral $APCD$ is equal to the area of the triangle $ABP$.