Example 1. $\mathrm{AC}$ and $\mathrm{CE}$ are two diagonals of a regular hexagon $\mathrm{ABCDEF}$, and points $\mathrm{M}$ and $\mathrm{N}$ internally divide $\mathrm{AC}$ and $\mathrm{CE}$, respectively, such that $\mathrm{M}: \mathrm{AC} = \mathrm{CN}: \mathrm{CE} = \mathrm{r}$. If points $\mathrm{B}$, $\mathrm{M}$, and $\mathrm{N}$ are collinear, find the value of $\mathrm{r}$. (1973 IMO Problem)
Connect $B D$, $N D$. Since $\triangle M B C$ rotates counterclockwise by $120^{\circ}$ around point $O$ and coincides with $\triangle \mathrm{NDE}$, we know that $\angle B N D=120^{\circ}$. Construct an equilateral triangle $B D G$ with $B D$ as one side towards the direction of point $N$, then $N$,
$$
\mathrm{CN}=\mathrm{CB} \text {. }
$$

Therefore, we can obtain
$$
\mathbf{r}=\mathrm{CN}: \mathrm{CE}=\mathrm{CB}: \mathrm{CE}=\frac{\sqrt{3}}{3} .
$$