例ミ As shown in the figure, $M$ and $N$ are internal points on the diagonals $AC$ and $CE$ of the regular hexagon $ABCDEF$, and $AM: AC = CN: CE = r$. If points $B$, $M$, and $N$ are collinear, find the value of $r$.

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Analysis Based on the necessary and sufficient conditions for vectors to be collinear to solve the problem.
Solution Let $\overrightarrow{A C}=2 a, \overrightarrow{A F}=2 b$, then $\overrightarrow{A M}=r \overrightarrow{A C}=2 r a, \overrightarrow{A B}=a-b$.
Therefore, $\overrightarrow{B M}=\overrightarrow{B A}+\overrightarrow{A M}=\boldsymbol{b}+(2 r-1) \boldsymbol{a}$
Also, $\overrightarrow{C E}=\frac{1}{2} \overrightarrow{C A}+\frac{3}{2} \overrightarrow{A F}=3 \boldsymbol{b}-\boldsymbol{a}$
Thus, $\overrightarrow{C N}=r \overrightarrow{C E}=3 r b-r a$
Therefore, $\overrightarrow{M N}=\overrightarrow{M C}+\overrightarrow{C N}=2(1-r) a+3 r \boldsymbol{b}-r \boldsymbol{a}=3 r \boldsymbol{b}+(2-3 r) \boldsymbol{a}$
Since $B, M, N$ are collinear
Therefore, $\frac{1}{3 r}=\frac{2 r-1}{2-3 r}$ Solving for $r$ gives $r=\frac{\sqrt{3}}{3}$