Example 8 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 points $F$ and $G$ on the extension of $AB$ such that $AF = AC$, $GF = AB$. Connect $FC$, $GC$. It is easy to see that $\triangle FAC$ is an equilateral triangle, $\triangle FCG \sim \triangle ABC$, $\triangle CGB \sim \triangle CDE$.

Since $E$ is the midpoint of $BC$, we have
$$
\begin{array}{l}
\frac{S_{\triangle BOC}}{S_{\triangle CDE}} = \left(\frac{BC}{EC}\right)^2 = 4. \\
\therefore S_{\triangle BCC} = 4 S_{\triangle CDE}. \\
\because \frac{\sqrt{3}}{4} AC^2 = S_{\triangle ACF} = 2 S_{\triangle ABC} + S_{\triangle BCX} \\
= 2 S_{\triangle ABC} + 4 S_{\triangle CDE} = 2 \left(S_{\triangle ABC} + 2 S_{\triangle CDE}\right) \\
= 2 \sqrt{3}, \\
\therefore AC^2 = 8 \Rightarrow AC = 2 \sqrt{2}.
\end{array}
$$