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Easy similarity in parallelogram

Source: VII Caucasus Mathematical Olympiad

March 13, 2022
geometryparallelogramnumber theory

Problem Statement

In parallelogram ABCDABCD, points EE and FF on segments ADAD and CDCD are such that ∠BCE=∠BAF\angle BCE=\angle BAF. Points KK and LL on segments ADAD and CDCD are such that AK=EDAK=ED and CL=FDCL=FD. Prove that ∠BKD=∠BLD\angle BKD=\angle BLD.
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