Question 19
There is given a convex quadrilateral $A B C D$. Prove that there exists a point $P$ inside the quadrilateral such that
$$
\angle P A B+\angle P D C=\angle P B C+\angle P A D=\angle P C D+\angle P B A=\angle P D A+\angle P C B=90^{\circ}
$$
if and only if the diagonals $A C$ and $B D$ are perpendicular.
$$
\angle P A B+\angle P D C=\angle P B C+\angle P A D=\angle P C D+\angle P B A=\angle P D A+\angle P C B=90^{\circ}
$$
if and only if the diagonals $A C$ and $B D$ are perpendicular.