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SHARYGIN 2023 Problem 5

    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} .

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