For example, $8 A C$ and $C E$ are two diagonals of the regular hexagon $A B C D E F$, and points $M$ and $N$ internally divide $A C$ and $C E$ respectively, such that $A M: A C = C N: C E = r$. If points $B$, $M$, and $N$ are collinear, find $r$.

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The above text translated into English, preserving the original text's line breaks and format.
Proof: As shown in Figure 8, let $AC = 2$, then $A(-1, 0), C(1, 0), B(0, -\frac{\sqrt{3}}{3}), E(0, \sqrt{3})$
$$
\begin{array}{l}
\because AM: AC \\
= CN: CE = r,
\end{array}
$$
$\therefore$ By the section formula, we get
$$
M(2r-1, 0), N(1-r, \sqrt{3}r).
$$

Also, $\because B, M, N$ are collinear,
thus $k_{BM} = k_{MN}$,

i.e., $\square$
$$
\begin{array}{l}
\frac{\frac{\sqrt{3}}{3}}{2r-1} = \frac{\sqrt{3}r}{1-r-(2r-1)}. \\
\therefore 1-3r^2 = 0, r = \frac{\sqrt{3}}{3}.
\end{array}
$$

Note: $r$ can also be found using $|BN| = |BM| + |MN|$.