10. In $\triangle A B C, A B=9, B C=8$ and $A C=7$. The bisector of $\angle A$ meets $B C$ at $D$. The circle passing through $A$ and touching $B C$ at $D$ cuts $A B$ and $A C$ at $M$ and $N$ respectively. Find $M N$.
在 $\triangle A B C$ 中, $A B=9, B C=8$ 及 $A C=7 \circ \angle A$ 的角平分線交 $B C$ 於 $D$ 。穿過 $A$且與 $B C$ 相切於 $D$ 的圓分別交 $A B$ 和 $A C$ 於 $M$ 和 $N \circ$ 求 $M N$ 。
10. 6
10. By the angle bisector theorem, $\frac{B D}{D C}=\frac{A B}{A C}=\frac{9}{7}$ and so $B D=8 \times \frac{9}{9+7}=\frac{9}{2}$. By the power chord theorem, $B D^{2}=B M \times B A$ and so $B M=\frac{9}{4}$.
Finally, $M N$ is parallel to $B C$ since
$$
\angle B D M=\angle M A D=\angle D A N=\angle D M N \text {. }
$$

Hence we have $\frac{A M}{A B}=\frac{M N}{B C}$, or $\frac{9-\frac{9}{4}}{9}=\frac{M N}{8}$, so that $M N=6$.