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Czech-Polish-Slovak 2003 Geometry

Source: Czech-Polish-Slovak 2003 Q2

April 28, 2013
geometrygeometric transformationreflectiongeometry unsolved

Problem Statement

In an acute-angled triangle ABCABC the angle at BB is greater than 45∘45^\circ. Points D,E,FD,E, F are the feet of the altitudes from A,B,CA,B,C respectively, and KK is the point on segment AFAF such that ∠DKF=∠KEF\angle DKF = \angle KEF. (a) Show that such a point KK always exists. (b) Prove that KD2=FD2+AF⋅BFKD^2 = FD^2 + AF \cdot BF.
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