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2016.15

Part of Ukrainian TYM Qualifying - geometry

Problems(1)

<BFC = 60^o wanted in triangle with <A=120^o

Source: 2016 XIX All-Ukrainian Tournament of Young Mathematicians named after M. Y. Yadrenko, Qualifying p15

5/6/2021
A non isosceles triangle ABCABC is given, in which ∠A=120o\angle A = 120^o. Let ALAL be its angle bisector, AKAK be it's median, drawn from vertex AA, point OO be the center of the circumcircle of this triangle, FF be the point of intersection of the lines OLOL and AKAK. Prove that ∠BFC=60o\angle BFC = 60^o.
geometryanglesUkrainian TYM