Notice that $D$ is the $\text{A - excenter touch point}$ in $\triangle ABC$, and perform a $\mathcal{I} \left (D, -\sqrt{DB\cdot DC} \right) $ inversion to finish.

Let $E$ and $F$ be the intersections of the angle bisectors of $\angle DXB$ and $\angle DXC$ with $(DXB)$ and $(DXC)$ respectively ($E, F \ne D$). Let $Y$ and $Z$ be the projections from $E$ and $F$ to $BC$ respectively. Let $I$ be the incenter of $\triangle ABC$ and $P$ be the foot from $I$ to $BC$. Note that $D$ is the $A$-extouch point of $\triangle ABC$. It is well known that $BD = CP$ and $CD = BP$. We see that $E$ and $F$ are the arc midpoints of arcs $\overarc{DB}$ and $\overarc{DC}$, so $EB = ED$ and $FC = FD$. Thus, $Y$ and $Z$ are the midpoints of line segments $BD$ and $CD$ respectively. Then $\angle EDY = \angle EDB = \angle EXB = \frac{1}{2}\angle AXB = \frac{1}{2}\angle ACB = \angle ICP$. Together with $\angle DYE = \angle CPI = 90^{\circ}$, we get $\triangle EDY \sim \triangle ICP$. Similarly, we can show that $\triangle FDZ \sim \triangle IBP$. So $EY = IP \cdot \frac{YD}{PC} = IP \cdot \frac{YD}{BD} = \frac{1}{2}IP$ and $FZ = IP \cdot \frac{DZ}{BP} = IP \cdot \frac{DZ}{CD} = \frac{1}{2}IP = EY$. Hence, $EFZY$ is a rectangle and $EF \parallel BC$. Then, $\angle FED = \angle EDB = \angle EBD$ and $\angle EFD = \angle FDC = \angle FCD$, so $EF$ is tangent to $(DXB)$ and $(DXC)$, and we are done.