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concyclic, <OAD+<OBC= <ODA + <OCB = 90^o (2021 Kyiv City MO Round2 10.4)

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February 14, 2021
geometryConcyclicangles

Problem Statement

Inside the quadrilateral ABCDABCD marked a point OO such that ∠OAD+∠OBC=∠ODA+∠OCB=90o\angle OAD+ \angle OBC = \angle ODA + \angle OCB = 90^o. Prove that the centers of the circumscribed circles around triangles OADOAD and OBCOBC as well as the midpoints of the sides ABAB and CDCD lie on one circle.
(Anton Trygub)
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