CAPS 2003 Problem 2

In an acute-angled triangle $ABC$ the angle at $B$ is greater than $45^\circ$ . Points $D,E, F$ are the feet of the altitudes from $A,B,C$ respectively, and $K$ is the point on segment $AF$ such that $\angle DKF = \angle KEF$ . (a) Show that such a point $K$ always exists. (b) Prove that $KD^2 = FD^2 + AF \cdot BF$ .