9. Answer: 224
Let the circle with centre $\mathrm{D}$ and radius $r$ touch the tangent lines $\mathrm{AC}, \mathrm{BA}$ produced and $\mathrm{BC}$ produced at the points $\mathrm{E}, \mathrm{F}$ and $\mathrm{G}$ respectively. Then $r=\mathrm{DE}=\mathrm{DF}=\mathrm{DF}$. Hence, triangles $\mathrm{BDF}$ and $\mathrm{BDG}$ are congruent, and hence $\angle \mathrm{ABD}=\angle \mathrm{CBD}=1 / 2 \angle \mathrm{ABC}$. We have $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{\theta^{2}+7^{2}-9^{2}}{2(8)(7)}=\frac{2}{7}$, and hence $\sin \frac{B}{2}=\sqrt{\frac{1-\cos B}{2}}=\sqrt{\frac{5}{14}}$.
To find $r$, we have
$$(\mathrm{ABD})+(\mathrm{BCD})-(\mathrm{ACD})=(\mathrm{ABC}),$$
where (ABD) denotes the area of triangle $\mathrm{ABD}$, etc.
Hence $\frac{1}{2} c r+\frac{1}{2} a r-\frac{1}{2} b r=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{8+9+7}{2}=12$.
Solving, we get $r=4 \sqrt{5}$. Considering, triangle $B D F$, we have $B D=\frac{r}{\sin \frac{4}{2}}=4 \sqrt{14}$. Thus, we have $\mathrm{BD}^{2}=224$.