9. For a regular hexagon $A B C D E F$, the diagonals $A C$ and $C E$ are internally divided by points $M$ and $N$ in the following ratios: $\frac{A M}{A C}=\frac{C M}{C E}=r$. If points $B, M, N$ are collinear, find the ratio $r$.
9. Use the letters of the points to represent the corresponding complex numbers. According to the section formula, we have
$$
M=(1-r) A+r C, M=(1-r) C+r E .
$$

Since $B, M, N$ are collinear, we have $\left|\begin{array}{lll}1 & 1 & 1 \\ B & M & N \\ \bar{B} & \bar{M} & \bar{N}\end{array}\right|=0$,
which means $\left|\begin{array}{lll}1 & (1-r)+r & (1-r)+r \\ B & (1-r) A+r C & (1-r) C+r E \\ B & (1-r) \bar{A}+r \bar{C} & (1-r) \bar{C}+r \bar{E}\end{array}\right|=0$,

Thus, we can obtain
$$
(1-r)^{2}\left|\begin{array}{ccc}
1 & 1 & 1 \\
B & A & C \\
\bar{B} & \bar{A} & \bar{C}
\end{array}\right|+(1-r) r\left|\begin{array}{ccc}
1 & 1 & 1 \\
B & A & E \\
\bar{B} & \bar{A} & \bar{E}
\end{array}\right|+r^{2}\left|\begin{array}{ccc}
1 & 1 & 1 \\
B & C & E \\
\bar{B} & \bar{C} & \bar{E}
\end{array}\right|=0 .
$$

Notice that $\triangle B A C$ and $\triangle B A E$ are in the same direction, while both are in the opposite direction to $\triangle B C E$. Also, we know $S_{\triangle B M E}=S_{\triangle B E C}=2 S_{\triangle A B C}$, thus, combining the above relationship, we have $(1-r)^{2}+2(1-r) r-2 r^{2}=0$,
which simplifies to $3 r^{2}-1=0$, solving for $r$ gives $r=\frac{\sqrt{3}}{3}$.