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Moldova TST triangle geometry

Source: Moldova TST 2014, First Day, Problem 3

March 4, 2014
geometrytrigonometrypower of a pointradical axisgeometry proposed

Problem Statement

Let △ABC\triangle ABC be an acute triangle and ADAD the bisector of the angle ∠BAC\angle BAC with D∈(BC)D\in(BC). Let EE and FF denote feet of perpendiculars from DD to ABAB and ACAC respectively. If BF∩CE=KBF\cap CE=K and ⊙AKE∩BF=L\odot AKE\cap BF=L prove that DL⊥BFDL\perp BF.
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