$證明：(1) \because \triangle \mathrm{ABC}, \triangle \mathrm{CDE}都為等邊三角形 $
$\therefore A C=B C, C D=C E$, $\angle \mathrm{ACB}=\angle \mathrm{ECD}=\angle \mathrm{B}=60^{\circ}$
$\therefore \angle \mathrm{ACB}-\angle \mathrm{ACD}=\angle \mathrm{ECD}-\angle \mathrm{ACD}$ $\therefore \angle B C D=\angle A C E$
$在\triangle \mathrm{BDC}和\triangle \mathrm{AEC}中$
$\left\{\begin{array}{l}B C=A C \\ \angle B C D=\angle A C E \\ C D=C E\end{array}\right.$
$\therefore \triangle \mathrm{BDC} \cong \triangle \mathrm{AEC}(\mathrm{SAS})$
$(2)∵△BDC≌△AEC$
$\therefore \angle \mathrm{EAC}=\angle \mathrm{B}=60^{\circ}$
$\therefore \angle \mathrm{EAC}=\angle \mathrm{ACB}$ $\therefore A E / / B C$
