olymmath

RU-ALLRUSSIAN 1978 Problem 2

    ABCD is a quadrilateral. M is a point inside it such that ABMD is a parallelogram. ∠CBM=\angle \mathrm{CBM} = ∠CDM\angle \mathrm{CDM} . Show that ∠ACD=∠BCM\angle \mathrm{ACD} = \angle \mathrm{BCM} .

    Dataset text NuminaMath olympiads. Quoted with source.

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