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\frac{AD}{DC}=\frac{BC}{2CE}

Source: China second round 2018 (B) Q2

June 22, 2019
geometry

Problem Statement

In triangle △ABC,AB=AC.\triangle ABC, AB=AC. Let DD be on segment ACAC and EE be a point on the extended line BCBC such that CC is located between BB and EE and ADDC=BC2CE\frac{AD}{DC}=\frac{BC}{2CE}. Let ω\omega be the circle with diameter AB,AB, and ω\omega intersects segment DEDE at F.F. Prove that B,C,F,DB,C,F,D are concyclic.
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