Example 3. In an equilateral $\triangle ABC$, take a point $D$ inside such that $DA = DB$; also take a point $E$ outside $\triangle ABC$ such that $\angle DBE = \angle DBC$, and $BE = BA$. Find $\angle BED$. (1992, Sichuan Province Junior High School Mathematics League)
Connect $D C$.
$$
\because A D=B D, A C=
$$
$B C, D C$ is common,
$$
\therefore \triangle A D C \cong \triangle B D C \text {. }
$$

Thus $\angle B C D=30^{\circ}$.
Also, $\angle D B E=\angle D B C$, $B E=A B=B C, B D$ is common, $\therefore \triangle B D E \cong \triangle B D C$. Therefore, $\angle B E D=\angle B C D=30^{\circ}$.