9. In $\triangle A B C$, $A B=7, B C=8, C A=9$, through the incenter $I$ of $\triangle A B C$ draw $D E / / B C$, intersecting $A B$ and $A C$ at points $D$ and $E$ respectively. Then the length of $D E$ is $\qquad$
9. $\frac{16}{3}$.

As shown in Figure 6, let the side lengths of $\triangle ABC$ be $a, b, c$, the radius of the incircle $\odot I$ be $r$, and the altitude from vertex $A$ to side $BC$ be $h_{a}$. Then,
$$
\frac{1}{2} a h_{a}=S_{\triangle A B C}=\frac{1}{2}(a+b+c) r .
$$

Therefore, $\frac{r}{h_{a}}=\frac{a}{a+b+c}$.
Since $\triangle ADE \sim \triangle ABC$, the corresponding segments are proportional. Thus,
$$
\frac{h_{a}-r}{h_{a}}=\frac{DE}{BC} .
$$

Hence, $DE=\frac{h_{a}-r}{h_{a}} \cdot a=\left(1-\frac{r}{h_{a}}\right) a$
$$
=\left(1-\frac{a}{a+b+c}\right) a=\frac{a(b+c)}{a+b+c} \text {. }
$$

Therefore, $DE=\frac{8 \times(7+9)}{8+7+9}=\frac{16}{3}$.