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prove perpendicularity with equal lengths and midpoint

Source: Sharygin 2023 - P5 (Grade-8)

March 4, 2023
geometryperpendicularSharygin Geometry OlympiadSharygin 2023

Problem Statement

Let ABCDABCD be a cyclic quadrilateral. Points EE and FF lie on the sides ADAD and CDCD in such a way that AE=BCAE = BC and AB=CFAB = CF. Let MM be the midpoint of EFEF. Prove that AMC=90\angle AMC = 90^{\circ}.