olymmath

BALTIC-WAY 2000 Problem 2

    Given an isosceles triangle ABCA B C with ∠A=90∘\angle A=90^{\circ}. Let MM be the midpoint of ABA B. The line passing through AA and perpendicular to CMC M intersects the side BCB C at PP. Prove that ∠AMC=∠BMP\angle A M C=\angle B M P.

    Dataset text NuminaMath olympiads. Quoted with source.

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