Through the point $ K $ lying outside the circle $ \omega $, the tangents are drawn $ KB $ and $ KD $ to this circle ($ B $ and $ D $ are tangency points) and a line intersecting a circle at points $ A $ and $ C $. The bisector of angle $ ABC $ intersects the segment $ AC $ at the point $ E $ and circle $ \omega $ at $ F $. Prove that $ \angle FDE = 90^\circ $.