9. Given that $D$ is a point on the side $BC$ of the equilateral $\triangle ABC$ with side length 1, the inradii of $\triangle ABD$ and $\triangle ACD$ are $r_{1}$ and $r_{2}$, respectively. If $r_{1}+r_{2}=\frac{\sqrt{3}}{5}$, then there are two points $D$ that satisfy this condition, denoted as $D_{1}$ and $D_{2}$. The distance between $D_{1}$ and $D_{2}$ is $\qquad$.
9. $\frac{\sqrt{6}}{5}$.

Let $B D=x$. By the cosine theorem, we get $A D=\sqrt{x^{2}-x+1}$.
On one hand, $S_{\triangle B D D}=\frac{1}{2} x \cdot \frac{\sqrt{3}}{2}$;
On the other hand,
$S_{\triangle A B D}=\frac{1}{2}\left(1+x+\sqrt{x^{2}-x+1}\right) r_{1}$.
Solving this, we get $r_{1}=\frac{\sqrt{3}}{6}\left(1+x-\sqrt{x^{2}-x+1}\right)$.
Similarly, $r_{2}=\frac{\sqrt{3}}{6}\left(2-x-\sqrt{x^{2}-x+1}\right)$.
Thus, $r_{1}+r_{2}=\frac{\sqrt{3}}{6}\left(3-2 \sqrt{x^{2}-x+1}\right)$.
When $x=\frac{1}{2}$, $r_{1}+r_{2}$ reaches its maximum value, and the maximum value is $\frac{\sqrt{3}-1}{2}$. Therefore, $\frac{\sqrt{3}}{6}<r_{1}+r_{2} \leqslant \frac{\sqrt{3}-1}{2}$.
Since $r_{1}+r_{2}=\frac{\sqrt{3}}{5}$, then $x^{2}-x+\frac{19}{100}=0$.
Let the two roots be $x_{1}$ and $x_{2}$. Thus,
$$
\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=\frac{\sqrt{6}}{5} \text {. }
$$