Points $K$ and $L$ are inner for $AB$ for an acute $\Delta ABC$, where $K$ is between $A$ and $L$. Let $P,Q$, and $H$ be the feet of the perpendiculars from $A$ to $CK$, from $B$ to $CL$, and from $C$ to $AB$, respectively. Point $M$ is the middle point of $AB$. If $PH\cap AC=X$ and $QH\cap BC=Y$, prove that points $H,P,M$, and $Q$ lie on one circle, if and only if the lines $AY,BX$, and $CH$ intersect in one point.