Example 14. If $A C, C E$ are two diagonals of a regular hexagon $A B C D E F$, and points $M, N$ internally divide $A C, C E$ such that $A M: A C=C N: C E=r$. If $B, M, N$ are collinear, find $r$. (23rd IMO Problem)
Let $A C, B E$ intersect at $K$. By the collinearity of $B, M, N$ and Menelaus' theorem, we have
$$
\begin{array}{l}
\frac{C N}{N E} \cdot \frac{E B}{B K} \cdot \frac{K M}{M C} \\
=1 .
\end{array}
$$

Assume the side length of the regular hexagon is 1. Then
$$
A C=C E=\sqrt{3} \text {. }
$$

And $\frac{C N}{N E}=\frac{C N}{C E-C N}=\frac{\frac{C N}{C E}}{1 - \frac{C N}{C E}}=\frac{r}{1-r}$,
also $\frac{E}{B} \frac{P}{Z}=\frac{\mathscr{Z}}{\frac{1}{2}}=4$,
and $\frac{K M}{M C}=\frac{A M-A K}{A C-A M}=\frac{\frac{A M}{A C}-\frac{A K}{A C}}{1-\frac{A M}{A C}}=\frac{r-\frac{1}{2}}{1-r}$.
Substituting the above three expressions into equation (*), and simplifying, we get
$$
3 r^{2}=1, r=\frac{1}{\sqrt{3}} \text {. }
$$