9. As shown in Figure 1, in $\triangle A B C$, $A M$ is a median, intersecting the incircle $\Gamma$ of $\triangle A B C$ at points $K$ and $L$. If
$$
A K: K L: L M=1: 2: 3 \text {, }
$$

find $B C: C A: A B$.
9. Let $A K=t, K L=2 t, L M=3 t$, and circle $\Gamma$ is tangent to $B C$, $C A$, and $A B$ at points $D$, $E$, and $F$ respectively. Then
$$
C D=C E=x .
$$

Thus, $M D=\sqrt{15} t, A E=A F=\sqrt{3} t$.
It is easy to see that $M$ is the midpoint of $B C$, so
$$
\begin{array}{l}
B M=M C=x+\sqrt{15} t, \\
B F=B D=x+2 \sqrt{15} t .
\end{array}
$$

By the property that the sum of the squares of the sides of a parallelogram equals the sum of the squares of its diagonals, we have
$$
\begin{array}{l}
(12 t)^{2}+4(x+\sqrt{15} t)^{2} \\
=2\left((x+\sqrt{3} t)^{2}+(x+2 \sqrt{15} t+\sqrt{3} t)^{2}\right) \\
\Rightarrow t=\frac{\sqrt{3}}{9-3 \sqrt{5}} x .
\end{array}
$$

Thus, $B C=\frac{18}{9-3 \sqrt{5}} x, C A=\frac{12-3 \sqrt{5}}{9-3 \sqrt{5}} x$,
$$
A B=\frac{12+3 \sqrt{5}}{9-3 \sqrt{5}} x .
$$

Therefore, $B C: C A: A B=6:(4-\sqrt{5}):(4+\sqrt{5})$.