12. As shown in Figure 5, in $\triangle A B C$, it is given that $A C=B C, \angle C=$ $20^{\circ}, D$ and $E$ are points on sides $B C$ and $A C$ respectively. If $\angle C A D=20^{\circ}$, $\angle C B E=30^{\circ}$, find the size of $\angle A D E$.
12. As shown in Figure 8, draw $D G / / B A$, intersecting $A C$ at point $G$, and connect $B G$ with $A D$ intersecting at point $H$. Then
$$
\begin{array}{l}
A D=B G, \\
A H=B H, \\
D H=G H .
\end{array}
$$

In $\triangle A B C$, since $A C=$ $B C, \angle C=20^{\circ}$, therefore,
$$
\angle C A B=\angle C B A=80^{\circ} \text {, }
$$

we have $\angle H A B=\angle C A B-\angle C A D=60^{\circ}$.
Thus, $\triangle A B H$ and $\triangle G D H$ are both equilateral triangles.
Also, in $\triangle A B E$, we have
$$
\angle E B A=\angle C B A-\angle C B E=80^{\circ}-30^{\circ}=50^{\circ} \text {, }
$$

which gives $\angle A E B=180^{\circ}-\angle C A B-\angle E B A=50^{\circ}$.
Thus, $\angle A E B=\angle E B A$, so $A B=A E$.
Therefore, $A E=A H$.
Draw $D F / / B E$, intersecting $A C$ at point $F$, then $\triangle G D F \backsim \triangle A B E$.
Thus, $F G=D G$.
Hence, $F G=D H$.
(1) + (2) gives
$$
A E+F G=A H+D H=A D \text {. }
$$

Also, in $\triangle A B C$, since $\angle C A D=\angle C=20^{\circ}$, we get $A D=C D$. And since $C D=C G$, we have
$$
A D=C G=C F+F G \text {. }
$$

Comparing (3) and (4), we get $A E=C F$.
In summary, $\triangle A E D \cong \triangle C F D$, so
$$
\angle A D E=\angle C D F \text {. }
$$

Since $\angle A F D$ is an exterior angle of $\triangle C F D$, $\angle A F D=\angle A E B=50^{\circ}$, thus,
$$
\angle C D F=\angle A F D-\angle C=30^{\circ} .
$$

Therefore, $\angle A D E=30^{\circ}$.