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OA=OB if <PAD = <ADP=< CBP =< PCB =< CPD

Source: 2024 Czech and Slovak Olympiad III A p2

May 18, 2024
equal anglesgeometry

Problem Statement

Let the interior point PP of the convex quadrilateral ABCDABCD be such that ∣∠PAD∣=∣∠ADP∣=∣∠CBP∣=∣∠PCB∣=∣∠CPD∣.|\angle PAD| = |\angle ADP| = |\angle CBP| = |\angle PCB| = |\angle CPD|. Let OO be the center of the circumcircle of the triangle CPDCPD. Prove that ∣OA∣=∣OB∣|OA| = |OB|.
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