Example 5 In $\triangle A B C$, $E$ is the midpoint of $B C$, and $D$ is on $A C$. If $\angle B A C=60^{\circ}, \angle A C B=20^{\circ}$, $\angle D E C=80^{\circ}, S_{\triangle A B C}+2 S_{\triangle C D E}=\sqrt{3}$, find the length of $A C$.
Solution: As shown in Figure 8, take point $F$ on the extension of $AB$ such that $AF = AC$, and connect $FC$. Take point $G$ on $AF$ such that $GF = AB$, and connect $GC$. It is known that $\triangle FAC$ is an equilateral triangle; $\triangle CFG \cong \triangle CAB$; $\triangle CGB \sim \triangle CED$. We have $S_{\triangle CRG} = S_{\triangle ABC}$, $S_{\triangle COB} = 4 S_{\triangle CDE}$. From $S_{\triangle ACF} = 2 S_{\triangle ABC} + 4 S_{\triangle CDE} = 2 \sqrt{3}$, we have
$$
\begin{array}{l}
\frac{\sqrt{3}}{4} AC^2 = 2 \sqrt{3} . \\
\therefore AC = 2 \sqrt{2} .
\end{array}
$$

Here, from the "part" $\triangle ABC$, we think of the "whole" equilateral $\triangle FAC$, making full use of multiple conclusions, the solution is ingenious.