如图, $\angle 1=\angle 2, A B=A E, A C=A D$. 求证: $B C=E D$.

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证明: $\because \angle 1=\angle 2$,

$\therefore \angle 1+\angle E A C=\angle 2+\angle E A C$,

即 $\angle B A C=\angle E A D$,

在 $\triangle A B C$ 和 $\triangle A E D$ 中,

$\left\{\begin{array}{c}A B=A E \\ \angle B A C=\angle E A D \\ A C=A D\end{array}\right.$

$\therefore \triangle A B C \cong \triangle A E D(\mathrm{SAS})$,

$\therefore B C=E D$.