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BALTIC-WAY 1990 Problem 6

    Let ABCDA B C D be a quadrangle, ∣AD∣=∣BC∣,∠A+∠B=120∘|A D|=|B C|, \angle A+\angle B=120^{\circ} and let PP be a point exterior to the quadrangle such that PP and AA lie at opposite sides of the line DCD C and the triangle DPCD P C is equilateral. Prove that the triangle APBA P B is also equilateral.

    Dataset text NuminaMath olympiads. Quoted with source.

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