Problem

Source: Kyiv mathematical festival 2013

Tags: geometry, perpendicular, Angle Bisectors, angle bisector, parallelogram



Let $ABCD$ be a parallelogram ($AB < BC$). The bisector of the angle $BAD$ intersects the side $BC$ at the point K; and the bisector of the angle $ADC$ intersects the diagonal $AC$ at the point $F$. Suppose that $KD \perp BC$. Prove that $KF \perp BD$.