MathDB
Equal segments in a parallelogram

Source: ARO Regional stage 2023 10.8

February 16, 2023
geometryparallelogramRussia

Problem Statement

The bisector of ∠BAD\angle BAD of a parallelogram ABCDABCD meets BCBC at KK. The point LL lies on ABAB such that AL=CKAL=CK. The lines AKAK and CLCL meet at MM. Let (ALM)(ALM) meet ADAD after DD at NN. Prove that ∠CNL=90o\angle CNL=90^{o}
Back to ProblemsView on AoPS