Example 7 In $\triangle A B C$, $A B=A C, \angle A=$ $20^{\circ}$, take a point $D$ on side $A B$ such that $A D=B C$. Find the degree measure of $\angle B D C$.
Solution: As shown in Figure 7, construct a regular $\triangle E B C$ inside $\triangle A B C$ with $B C$ as a side, and connect $A E$. It is easy to see that $A E$ is the perpendicular bisector of $B C$, so
$$
\angle E A C=10^{\circ} \text {. }
$$

In $\triangle A D C$ and $\triangle C E A$,
$$
\begin{array}{c}
A D=B C=E C, A C=C A, \\
\angle D A C=20^{\circ}=\angle E C A . \\
\therefore \triangle A D C \cong \triangle C E A . \\
\text { Therefore, } \angle D C A=\angle E A C=10^{\circ} . \\
\therefore \angle B D C=\angle D A C+\angle D C A=30^{\circ} .
\end{array}
$$