Given a quadrangular pyramid $SABCD$, the basis of which is a convex quadrilateral $ABCD$. It is known that the pyramid can be tangent to a sphere. Let $P$ be the point of contact of this sphere with the base $ABCD$. Prove that $\angle APB + \angle CPD = 180^o$.
Problem
Source: Champions Tournament (Ukraine) - Турнір чемпіонів - 2008 Seniors p4
Tags: geometry, 3D geometry, sphere, pyramid, Champions Tournament