Given $A C, C E$ are two diagonals of a regular hexagon $A B C D E F$, points $M, N$ internally divide $A C, C E$ respectively, such that
$A M \cdot A C=C N: C E=$
$r$. If $B, M, N$ are collinear, find the value of $r$. (23rd IMO)
Solve: Since $B, M, N=$ are collinear points, then we have
$$
S_{\triangle B C C^{2}}=S_{\triangle B C M}+S_{\triangle C M .} .
$$

Let the side length of the equilateral triangle be $a$, then $A C=C E=\sqrt{3} a$.
$$
\begin{array}{l}
\text { Also, } A M: A C=C N: C E=r, \\
\therefore A M=C N=\sqrt{3} a r . \\
\text { Also, } \because \angle B C D=120^{\circ}, \angle A C B=30^{\circ}, \\
\therefore \angle B C E=\angle A C D=90^{\circ}, \angle A C E=60^{\circ} .
\end{array}
$$

Substituting into (1) we get
$$
\begin{array}{l}
\frac{1}{2} a(\sqrt{3} a-\sqrt{3} a r) \sin 30^{\circ}+\frac{1}{2} \sqrt{3} a r \\
-(\sqrt{3} a-\sqrt{3} a r) \sin 60^{\circ} \\
=\frac{1}{2} \sqrt{3} a^{2} r . \\
\therefore 3 r^{2}-1=0 .
\end{array}
$$

Solving, we get $r=\frac{\sqrt{3}}{3}, r^{\prime}=-\frac{\sqrt{3}}{3}$ (discard the negative value).