In triangle $ABC, \angle A = 90^o$. $M$ is the midpoint of $BC$ and $H$ is the foot of the altitude from $A$ to $BC$. The line passing through $M$ and perpendicular to $AC$ meets the circumcircle of triangle $AMC$ again at $P$. If $BP$ intersects $AH$ at $K$, prove that $AK = KH$.