The image of $D$ under inversion with pole $H$ which sends circumcirlce of $ABC$ to nine-point circle is that common point !

If $BA=BC$, then $DH$ passes through $B$. Done. If $BA\neq BC$, then let the ray $DH$ cuts the circumcircle of $ABC$ at $K$. If we can show that $\angle BKH=90^\circ$, then we are done (Because this indicates $K$ lies on the circle with diameter $[BH]$). We know that the symmetric of $H$ with respect to $D$, say $L$, lies on the circumcircle of $ABC$ and $BL$ is the diameter of the circumcircle of $ABC$. Hence, $\angle BKH=\angle BKL=90^\circ$, as desired.

Let $B’ $ be the diametrically opposite of $B$. Let $DH$ meets $(ABC)$ at $P$. Since $B’,D,H$ are collinear, $\angle{BPH}=\angle{BPB’}=90^\circ$. Which implies $P\in(BH)$