$(1)解:(1)\ \because A C=B C,$
$\therefore \angle A B C=\angle B A C, $
$\because C E=C D, $
$\therefore \angle C D E=\angle C E D, $
$又 \because \angle A B C=\angle C D E, $
$\therefore \angle A B C=\angle B A C=\angle C D E=\angle C E D, $
$\because \angle A C B=180^{\circ}-\angle A B C-\angle B A C, $
$\angle E C D=180^{\circ}-\angle C D E-\angle C E D,$
$\therefore \angle A C B=\angle D C E,$
$\therefore \angle A C B-\angle A C D=\angle E C D-\angle A C D,$
$\therefore \angle B C D=\angle A C E,$
$在 \triangle A E C 和 \triangle B D C 中,$
${{\begin{cases}{{A C=B C}}\\{\angle A C E=\angle B C D}\\{\ C E=C D}\end{cases}}}$
$\therefore \triangle A E C \cong \triangle B D C(\mathrm {SAS})$
$\therefore A E=B D$