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Prove that <EBF=<ECF

Source: Dutch IMO TST I Problem 4

July 17, 2014
geometrygeometric transformationreflectionangle bisectorgeometry proposed

Problem Statement

Let △ABC\triangle ABC be a triangle with ∣AC∣=2∣AB∣|AC|=2|AB| and let OO be its circumcenter. Let DD be the intersection of the bisector of ∠A\angle A with BCBC. Let EE be the orthogonal projection of OO to ADAD and let F≠DF\ne D be the point on ADAD satisfying ∣CD∣=∣CF∣|CD|=|CF|. Prove that ∠EBF=∠ECF\angle EBF=\angle ECF.
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