Example 13 (adapted from Example 12) As shown in Figure 2.5.13, AC and CE are two diagonals of the regular hexagon $A B C D E F$. Points $M$ and $N$ internally divide $A C$ and $C E$ respectively, such that $\frac{A M}{A C}=\frac{C N}{C E}=r$. If points $B, M$, and $N$ are collinear, find $r$.
Given that $B, M, N$ are collinear, then
$$
\frac{S_{\triangle C E A}}{\frac{C B}{C B}}+\frac{S_{\triangle A B C}}{\frac{C N}{C E}}=\frac{S_{\triangle B C E}}{\frac{C M}{C A}}
$$

If we set $A B=1$, then $S_{\triangle C E A}=\frac{3}{4} \sqrt{3}, S_{\triangle A B C}=\frac{\sqrt{3}}{4}, S_{\triangle B C E}=\frac{\sqrt{3}}{2}$.
$$
\text { Also, } \frac{C M}{C A}=1-r \text {, thus } \frac{3}{4} \sqrt{3}+\frac{\frac{\sqrt{3}}{4}}{r}=\frac{\frac{\sqrt{3}}{2}}{1-r} \text {. }
$$

Therefore, solving for $r$ yields $r=\frac{\sqrt{3}}{3}$ (negative value discarded).