Školjka

%V0 There is given a convex quadrilateral $ABCD$. Prove that there exists a point $P$ inside the quadrilateral such that $$\angle PAB + \angle PDC = \angle PBC + \angle PAD = \angle PCD + \angle PBA = \angle PDA + \angle PCB = 90^{\circ}$$ if and only if the diagonals $AC$ and $BD$ are perpendicular. Proposed by Dukan Dukic, Serbia