olymmath

TSTST 2022 Problem 7

    Let ABCDA B C D be a parallelogram. Point EE lies on segment CDC D such that 2∠AEB=∠ADB+∠ACB,2 \angle A E B=\angle A D B+\angle A C B, and point FF lies on segment BCB C such that 2∠DFA=∠DCA+∠DBA.2 \angle D F A=\angle D C A+\angle D B A . Let KK be the circumcenter of triangle ABDA B D. Prove that KE=KFK E=K F.

    Dataset text NuminaMath olympiads. Quoted with source.

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