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SHARYGIN 2016 Problem 1

    A trapezoid ABCDABCD with bases ADAD and BCBC is such that AB=BDAB = BD. Let MM be the midpoint of DCDC. Prove that ∠MBC=∠BCA\angle MBC = \angle BCA.

    Official text Sharygin Geometry Olympiad 2016 official paper, page 2. Quoted with source.

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