Example 6 In $\triangle A B C$, it is known that $A C=B C, \angle C=$ $20^{\circ}, D 、 E$ are points on sides $B C 、 A C$ respectively. If $\angle C A D$ $=20^{\circ}, \angle C B E=30^{\circ}$, find the degree measure of $\angle A D E$.
Solve as shown in Figure 8, take point $F$ on $BC$ such that $AF=AB$.
It is easy to know that $\angle ABC=80^{\circ}$.
Notice that
$$
\begin{array}{l}
\angle AEB \\
=\angle ACB + \angle EBC \\
=50^{\circ} \\
=\angle ABC - \angle EBC \\
=\angle ABE. \\
\text{Therefore, } AB=AE.
\end{array}
$$

Thus, $A$ is the circumcenter of $\triangle BEF$.
Then $\angle EAF = 2 \angle EBC = 60^{\circ}$.
So, $\triangle AEF$ is an equilateral triangle.
Therefore, $AF=EF$, $\angle AFE=60^{\circ}$.
In $\triangle ADF$, we have
$$
\begin{array}{l}
\angle DAF = \angle EAF - \angle CAD = 40^{\circ}, \\
\angle ADF = \angle C + \angle CAD = 40^{\circ}. \\
\text{Hence, } \angle DAF = \angle ADF. \\
\text{Therefore, } AF = DF.
\end{array}
$$

Thus, $AF=EF=DF$, which means $F$ is the circumcenter of $\triangle AED$.
By property 2, we know
$$
\angle ADE = \frac{1}{2} \angle AFE = 30^{\circ}.
$$