Example 4.4.5 (IMO 23) In a regular hexagon $A B C D E F$, the diagonals $A C$ and $C E$ are divided by internal points $M$ and $N$ in the following ratios: $A M: A C = C N: C E = r$. If points $B$, $M$, and $N$ are collinear, find the ratio $r$.

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In the plane, establish a complex plane, let the coordinates of points $A, B, C, \cdots$ be $A, B, C, \cdots$ respectively. According to the section formula, we have
$$
M=(1-r) A+r C, N=(1-r) C+r E .
$$

Since $B, M, N$ are collinear, i.e., the area of $\triangle B M N$ is 0, we have
$$
\left|\begin{array}{ccc}
1 & 1 & 1 \\
B & M & N \\
\bar{B} & \bar{M} & \bar{N}
\end{array}\right|=0,
$$

which is $\left|\begin{array}{ccc}1 & (1-r)+r & (1-r)+r \\ B & (1-r) A+r C & (1-r) C+r E \\ \bar{B} & (1-r) \bar{A}+r \bar{C} & (1-r) \bar{C}+r \bar{E}\end{array}\right|=0$. By the properties of determinants, we get
$$
(1-r)^{2}\left|\begin{array}{ccc}
1 & 1 & 1 \\
B & A & C \\
\bar{B} & \bar{A} & \bar{C}
\end{array}\right|+(1-r) r\left|\begin{array}{ccc}
1 & 1 & 1 \\
B & A & E \\
\bar{B} & \bar{A} & \bar{E}
\end{array}\right|+r^{2}\left|\begin{array}{ccc}
1 & 1 & 1 \\
B & C & E \\
\bar{B} & \bar{C} & \bar{E}
\end{array}\right|=0 \text {. }
$$

Notice that, triangles $B A C, B A E$ are in the same direction, and they are in the opposite direction to triangle $B C E$;
Secondly, triangle $B A E$ has the same area as triangle $B C E$. This area is twice the area of triangle $B A C$. Therefore, from the above equation, we get: $(1-r)^{2}+2(1-r) r-2 r^{2}=0$.
Simplifying, we get $3 r^{2}-1=0$, from which we solve $r=\sqrt{3} / 3$.