Solve As shown in Figure 1.4.10, extend the original figure to form a regular $\triangle A C F$ with $A C$ as a side. It is easy to see that point $B$ must lie on $A F$. Construct the angle bisector of $\angle B C F$ intersecting $A F$ at $G$, then
$$
\angle F C G=\angle G C B=\angle A C B=20^{\circ}
$$

and $\angle F G C=100^{\circ}=\angle A B C$, thus
$$
\angle C B G=\angle C G B=\angle D E C=80^{\circ}
$$

Therefore, $\triangle C G B \sim \triangle C D E$, and the similarity ratio is 2, i.e.,
$$
S_{\triangle C B G}=4 S_{\triangle C D E}
$$

Hence,
$$
2 S_{\triangle A B C}+4 S_{\triangle C D E}=S_{\triangle A C F}=\frac{\sqrt{3}}{4}
$$

Thus, the required value is $\frac{\sqrt{3}}{8}$.
Example 9 In $\triangle A B C$, it is known that point $E$ is the midpoint of side $B C$, $D$ is on $A C$, $A C=1, \angle B A C=$ $60^{\circ}, \angle A B C=10^{\circ}, \angle D E C=80^{\circ}$, find the value of $S_{\triangle A B C}+2 S_{\triangle C D E}$.