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Feet of perpendiculars - Switzerland 2011

Source:

March 22, 2011
geometryincentertrigonometrytrig identitiesLaw of Sinesgeometry proposed

Problem Statement

Let △ABC\triangle{ABC} be an acute-angled triangle and let DD, EE, FF be points on BCBC, CACA, ABAB, respectively, such that \angle{AFE}=\angle{BFD}\mbox{,} \angle{BDF}=\angle{CDE} \mbox{and} \angle{CED}=\angle{AEF}\mbox{.} Prove that DD, EE and FF are the feet of the perpendiculars through AA, BB and CC on BCBC, CACA and ABAB, respectively.
(Swiss Mathematical Olympiad 2011, Final round, problem 2)
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