In the acute-angled triangle $ ABC $, the sides $ AB $ and $BC$ have different lengths, and the extension of the median $ BM $ intersects the circumscribed circle at the point $ N $. On this circle we note such a point $ D $ that $ \angle BDH = 90 {} ^ \circ $, where $ H $ is the point of intersection of the altitudes of the triangle $ ABC $. The point $K$ is chosen so that $ ANCK $ is a parallelogram. Prove that the lines $ AC $, $ KH $ and $ BD $ intersect at one point. (Igor Nagel)