Let $F$ be the midpoint of $DH$, $K$ be the midpoint of $AH$, $L$ on $BC$ the foot of the altitude from $A$, and let $M$ be the intersection point of $BC$ and $DE$.

Because $AE$ is a diameter of the circumcircle, $\angle ACE = \angle ABE = 90^\circ$, so $CE$ is perpendicular to $AC$ and $BE$ is perpendicular to $AB$. Because $BH$ is on the altitude which is perpendicular to $AC$, we get $BH \parallel CE$. Similarly, $CH$ is perpendicular to $AB$ and so $CH \parallel BE$. This shows that $CHBE$ is a parallelogram and $M$ is the intersection point of its diagonals. Hence $M$ is the midpoint of $BC$ and thus is on the nine point circle.

Because $AE$ is a diameter of the circle, $\angle ADE = 90^\circ$. As $F, K$ are the midpoints of $DH$ and $AH$, respectively, we obtain $\angle KFM = 90^\circ$. Because $\angle KLM = 90^\circ$ as well, the quadrilateral $KFLM$ is cyclic.

As $K, L$ and $M$ are on the nine point circle, $F$ is on that circle as well.