Example 2. In parallelogram $A B C D$, it is known that $B C=$ $\sqrt{6} \mathrm{~cm}, A C=(3+$ $\sqrt{3}) \mathrm{cm}, A B=2 \sqrt{3}$ $\mathrm{cm}$. If the paper is folded so that point $A$ coincides with point $C$. Find the length of the fold.
Fold line $E S$ is the perpendicular bisector of $A C$.
$$
\begin{aligned}
\because \cos \angle C A B & =\frac{A B^{2}+A C^{2}-B C^{2}}{2 A B \cdot A C} \\
& =\frac{\sqrt{3}}{2},
\end{aligned}
$$
$$
\therefore \angle C A B=30^{\circ} \text {. }
$$
$\because A B C D$ is a $\square$,
$$
\therefore \angle D C A=\angle C A B=30^{\circ} \text {. }
$$

Therefore, $E F=2 F O$
$$
\begin{array}{l}
=2 \cdot O C \cdot \operatorname{tg} \angle D C A \\
=2 \cdot \frac{3+\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{3} \\
=1+\sqrt{3}(\mathrm{~cm}) .
\end{array}
$$