Let △ABC be an acute-angled triangle and let D, E, F be points on BC, CA, AB, respectively, such that \angle{AFE}=\angle{BFD}\mbox{,} \angle{BDF}=\angle{CDE} \mbox{and} \angle{CED}=\angle{AEF}\mbox{.} Prove that D, E and F are the feet of the perpendiculars through A, B and C on BC, CA and AB, respectively.(Swiss Mathematical Olympiad 2011, Final round, problem 2) geometryincentertrigonometrytrig identitiesLaw of Sinesgeometry proposed