Example 1 As shown in Figure $1, M$ and $N$ are the internal division points of the diagonals $A C$ and $C E$ of the regular hexagon $A B C D E F$, and $\frac{A M}{A C}=$
$\frac{C N}{C E}=\lambda$. If points $B$, $M$, and $N$ are collinear, find the value of $\lambda$.
(23rd IMO)
Solution: Extend $E A$ and $C B$ to intersect at point $P$. Let the side length of the regular hexagon be 1, then $P B=2$, and $A$ is the midpoint of $E P$, so $E A=A P$ $=\sqrt{3}$.
Since $\boldsymbol{A M}=\lambda A C$, then $C M=(1-\lambda) C A$.
Since $C P=3 C B$, and $C A$ is the median of $\triangle P C E$ on side $P E$, $C N=\lambda C E$, so,
$$
C A=\frac{1}{2}(C E+C P)=\frac{1}{2 \lambda} C N+\frac{3}{2} C B .
$$

Also, $C A=\frac{C M}{1-\lambda}$, hence
$$
\begin{array}{l}
\frac{1}{2 \lambda} C N+\frac{3}{2} C B=\frac{C M}{1-\lambda}, \\
C M=\frac{1-\lambda}{2 \lambda} C N+\frac{3(1-\lambda)}{2} C B .
\end{array}
$$

When $B$, $M$, and $N$ are collinear, there exists a real number $t$ such that $C M=(1-t) C N+t C B$. Therefore,
$$
\frac{1-\lambda}{2 \lambda}+\frac{3(1-\lambda)}{2}=1 \text {. }
$$

Solving for $\lambda$ gives $\lambda=\frac{\sqrt{3}}{3}$.