$ 解:(1)线段AB,CE的关系为AB=CE,AB⊥CE$
$ 理由:延长CE交AB于点F$
$ \because \angle A D C=90^{\circ}, \angle D A C=45^{\circ}$
$ \therefore \angle A C D=\angle D A C=45^{\circ}\therefore A D=C D$
$ 在 \triangle A D B 和 \triangle C D E 中$
$ \begin{cases}A D=C D\\\angle A D B=\angle C D E\\D B=D E\end{cases}$
$ \therefore \triangle A D B ≌\triangle C D E(\mathrm {SAS})$
$ \therefore A B=C E, \angle B A D=\angle D C E $
$ \because \angle B A D+\angle A B D=90^{\circ}$
$ \therefore \angle D C E+\angle A B D=90^{\circ}$
$\therefore \angle B F C=90^{\circ}$
$ \therefore A B \perp C E$