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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

Slični zadaci

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