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E satisfies AE/ED=CD/DB

Source: Baltic Way 1998

January 11, 2011
geometry proposedgeometry

Problem Statement

Given acute triangle ABCABC. Point DD is the foot of the perpendicular from AA to BCBC. Point EE lies on the segment ADAD and satisfies the equation AEED=CDDB\frac{AE}{ED}=\frac{CD}{DB} Point FF is the foot of the perpendicular from DD to BEBE. Prove that ∠AFC=90∘\angle AFC=90^{\circ}.
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