Let $O$ be the circumcenter of $ABC$. Since $N$ is the midpoint of $AH$, we want to prove that $D$ is on the Thales circle around $N$ which is equivalent to $\angle DAN=\angle NDA$. But since $\angle BAO=\angle HAC=90^\circ-\gamma$, $AD$ is also the angular bisector of $OAH$. Hence it suffices to prove that $\angle NDA=\angle OAD$ i.e. that $OA$ and $MN$ are parallel. But since clearly $AN \parallel OM$ (both are perpendicular to $BC$) this is equivalent to $OAMN$ being a parallelogram i.e. it suffices to check that $AN$ and $OM$ have the same length i.e. that $OM$ is half as long as $AH$. Let $X$ be the point of reflection of $H$ by $M$. It is well-known that $X$ lies on the circumcircle. Moreover, we have $HC \parallel BX$ so $\angle XBA=90^\circ$ so $X$ is opposite to $A$ on the circumcircle. But then $O,X,A$ are collinear so triangles $AHX$ and $OMX$ are similar whence finally $\frac{AH}{OM}=\frac{AX}{OX}=2$.

Let $BE,CF$ altitudes of $\triangle ABC$. Clearly $EF$ is radical axis of the circles of diameters $(AH), (BC)$, thus $EF\bot MN$. Since $MN$ is perpendicular bisector of $EF$, we get its intersection with the bisector of $\angle BAC$ lies onto the circle $\odot (AEF)$. Best regards, sunken rock