1. 如图,在等腰直角 $\triangle ABC$ 中,$AC = BC$,$\angle ACB = 90^{\circ}$,$D$ 为 $\triangle ABC$ 外一点,连接 $AD$,$CD$,$BD$,$\angle ADC = 45^{\circ}$. 求证:$AD\perp BD$.

证明:过点 C 作 $ CE \perp CD $ 交 DA 的延长线于点 E,
$ \therefore \angle DCE = 90 ^ { \circ } $.
$ \because \angle ACB = 90 ^ { \circ } $,
$ \therefore \angle ACE = \angle BCD $.
$ \because \angle ADC = 45 ^ { \circ } $,
$ \therefore CE = DC $.
$ \because AC = BC $,
$ \therefore \triangle ACE \cong \triangle BCD ( SAS ) $,
$ \therefore \angle E = \angle CDB = 45 ^ { \circ } $,
$ \therefore \angle ADB = \angle ADC + \angle CDB = 90 ^ { \circ } $,
$ \therefore AD \perp BD $.

2. 如图,$\triangle ABC$ 为等腰直角三角形,$\angle BAC = 90^{\circ}$,$AB = AC$,$D$ 是 $AC$ 边上一点.
(1) 如图 1,若 $CE\perp BD$ 交 $BD$ 的延长线于点 $E$,连接 $AE$,求证:$\angle AEB = 45^{\circ}$;

(2) 如图 2,若 $\angle AEB = 45^{\circ}$,求证:$CE\perp BD$;

(3) 如图 3,若 $\angle AEC = 135^{\circ}$,求证:$CE\perp BD$.

(1) 证明:在 BE 上截取 $ BF = CE $,连接 AF.
$ \because CE \perp BD $,
$ \therefore \angle CED = \angle BAC = 90 ^ { \circ } $,
$ \therefore \angle ACE + \angle CDE = \angle ABF + \angle ADB = 90 ^ { \circ } $.
又 $ \because \angle ADB = \angle CDE $,
$ \therefore \angle ABF = \angle ACE $.
又 $ \because AB = AC $,
$ \triangle ABF \cong \triangle ACE ( SAS ) $,
$ \therefore AF = AE $, $ \angle BAF = \angle CAE $,
$ \therefore \angle BAF + \angle CAF = \angle CAF + \angle CAE = 90 ^ { \circ } $,
$ \therefore \angle EAF = 90 ^ { \circ } $.
又 $ \because AF = AE $,
$ \therefore \triangle AEF $ 为等腰直角三角形,
$ \therefore \angle AEB = 45 ^ { \circ } $;
(2) 证明:过点 A 作 $ AF \perp AE $ 交 BE 于点 F.
$ \because \angle AEB = 45 ^ { \circ } $,
$ \therefore \triangle AFE $ 为等腰直角三角形,
$ \therefore AF = AE $, $ \angle AFE = \angle AEB = 45 ^ { \circ } $.
$ \because \angle BAF + \angle CAF = \angle CAE + \angle CAF = 90 ^ { \circ } $,
$ \therefore \angle BAF = \angle CAE $.
$ \because AB = AC $,
$ \therefore \triangle BAF \cong \triangle CAE ( SAS ) $,
$ \therefore \angle BFA = \angle CEA = 135 ^ { \circ } $,
$ \therefore \angle CED = 90 ^ { \circ } $,
$ \therefore CE \perp BD $;
(3) 证明:过点 A 向上作 $ AF \perp AE $ 交 CE 的延长线于点 F.
$ \because \angle AEC = 135 ^ { \circ } $,
$ \therefore \angle AEF = 45 ^ { \circ } $,
$ \therefore \triangle AEF $ 为等腰直角三角形,
$ \therefore AE = AF $.
$ \because \angle BAC = \angle EAF = 90 ^ { \circ } $,
$ \therefore \angle BAC + \angle CAE = \angle CAE + \angle EAF $,
$ \therefore \angle BAE = \angle CAF $.
$ \because AB = AC $, $ AE = AF $,
$ \therefore \triangle BAE \cong \triangle CAF ( SAS ) $,
$ \therefore \angle BEA = \angle F = 45 ^ { \circ } $,
$ \therefore \angle CEB = \angle AEC - \angle AEB = 90 ^ { \circ } $,
$ \therefore CE \perp BD $.