1. 如图,$\triangle ABC$中,$\angle ABC = 45^{\circ}$,$D是\triangle ABC$外一点,$DC \perp AC$,连接$BD$。当$\angle DBC = 45^{\circ}$时,求证：$DC = AC$。

如图，过点$C$作$CE\perp AB$于点$E$，作$CF\perp BD$交$BD$的延长线于点$F$．$\because \angle ABC=\angle DBC=45^{\circ}$，$\therefore BC$为$\angle ABD$的平分线，$\therefore CE=CF$．在四边形$BECF$中，$\because \angle EBF=\angle ABC+\angle OBC=90^{\circ}$，$\angle BEC=90^{\circ}$，$\angle BFC=90^{\circ}$，$\therefore \angle ECF=90^{\circ}$．$\because \angle ACD=90^{\circ}$，$\therefore \angle ACE+\angle ECD=\angle FCD+\angle ECD$，即$\angle ACE=\angle FCD$．$\because CE=CF$，$\angle CEA=\angle CFD=90^{\circ}$，$\therefore \triangle ACE\cong \triangle DCF(ASA)$，$\therefore DC=AC$．

2. 如图,$\triangle ABC$为等腰直角三角形,$\angle BAC = 90^{\circ}$,$AB = AC$,$D是AC$上一点。
(1) 若$\angle AEB = 45^{\circ}$,求证：$CE \perp BD$；
(2) 若$\angle AEC = 135^{\circ}$,求证：$CE \perp BD$。

(1)如图①，过点$A$作$AF\perp AE$交$BE$于点$F$．$\because \angle AEB=45^{\circ}$，$\therefore \angle AFE=180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}$，$\therefore \triangle AEF$是等腰直角三角形，易证得$\triangle ABF\cong \triangle ACE(SAS)$，$\therefore \angle ABE=\angle ACE$．又$\because \angle ADB=\angle EDC$，$\angle ABE+\angle ADB=90^{\circ}$，$\therefore \angle ACE+\angle EDC=90^{\circ}$，$\therefore \angle BEC=90^{\circ}$，$\therefore CE\perp BD$．
(2)如图②，过点$A$作$AF\perp AE$交$CE$的延长线于点$F$．$\because \angle AEC=135^{\circ}$，$\therefore \angle AEF=45^{\circ}$，$\therefore \angle AFE=180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}$，$\therefore \triangle AEF$是等腰直角三角形，易证得$\triangle ABE\cong \triangle ACF(SAS)$，$\therefore \angle ABE=\angle ACE$．又$\because \angle ADB=\angle EDC$，$\angle ABE+\angle ADB=90^{\circ}$，$\therefore \angle ACE+\angle EDC=90^{\circ}$，$\therefore \angle BEC=90^{\circ}$，$\therefore CE\perp BD$．
技法点拨
模型：已知$AB=AC$，$\angle BAC=90^{\circ}$，$\angle APB=45^{\circ}$，解题策略是过点$A$作$AD\perp AP$，构造等腰直角三角形$ADP$，结论是$\triangle ABD\cong \triangle ACP$，$CP\perp BP$；已知$AB=AC$，$\angle BAC=90^{\circ}$，$\angle APC=135^{\circ}$，解题策略是过点$A$作$AD\perp AP$，构造等腰直角三角形$ADP$，结论是$\triangle ABP\cong \triangle ACD$，$CP\perp BP$；已知$AB=AC$，$\angle BAC=90^{\circ}$，$\angle APC=45^{\circ}$，解题策略是过点$A$作$AD\perp AP$，构造等腰直角三角形$ADP$，结论是$\triangle ABD\cong \triangle ACP$，$PD\perp BD$。

3. 如图①,$\triangle ABC和\triangle DCE$都是等腰直角三角形,点$D在AC$上,连接$BE$,取$BE的中点M$,连接$AM$,$DM$。
(1)$AM与DM$具有怎样的位置关系和数量关系？请说明理由。经过探究,小李得到了一种解题思路：如图②,延长$DM交AB于点N$,利用“三线合一”可证$AM \perp DM$,且$AM = DM$。请利用小李的思路将完整的解题过程写出来。
(2) 如图③,若将$\triangle DCE绕点C逆时针旋转45^{\circ}$,即恰好使$AC \perp CE$,$DC平分\angle ACE$。则$AM与DM$具有怎样的位置关系和数量关系？请说明理由。
(3) 若将$\triangle DCE绕点C$逆时针旋转任意角度,如图④,则$AM与DM$具有怎样的位置关系和数量关系？(直接写出结果)

(1)$\because \triangle ABC$和$\triangle DCE$都是等腰直角三角形，$\therefore DE// AB$．又$\because M$为$BE$的中点，$\therefore$易证$\triangle MDE\cong \triangle MNB(ASA)$，$\therefore DE=BN=DC$，$DM=MN$．$\because \triangle ABC$为等腰直角三角形，$\therefore AC=AB$，$\therefore AN=AD$．在等腰直角$\triangle ADN$中，$M$为$DN$的中点，$\therefore AM\perp DM$，$AM=DM$．
(2)$AM\perp DM$，$AM=DM$．理由：延长$DM$交$BC$于$N$，连接$AN$，$AD$．如图①，$\because \angle DEC+\angle BCE=180^{\circ}$，$\therefore DE// BC$，$\therefore$易证$\triangle MDE\cong \triangle MNB(ASA)$，$\therefore BN=DE=DC$，$DM=MN$，$\therefore \triangle ABN\cong \triangle ACD(SAS)$，$\therefore \angle BAN=\angle CAD$，$AN=AD$．$\because \angle BAN+\angle CAN=90^{\circ}$，则$\angle CAD+\angle CAN=90^{\circ}$，$\therefore \triangle ADN$为等腰直角三角形．在等腰直角三角形$ADN$中，$M$为$DN$的中点，$\therefore AM\perp DM$，$AM=DM$．
(3)$AM\perp DM$，$AM=DM$．解析：过点$B$作$BN// DE$，交$DM$的延长线于点$N$，连接$AN$，$AD$．如图②，$\therefore$易证$\triangle MDE\cong \triangle MNB(ASA)$，$\therefore \angle BNM=\angle EDM$，$DM=MN$，$BN=DE=DC$．记$AC$与$DM$的交点为$F$，在四边形$ABNF$和四边形$CEDF$中，$\angle BNF=\angle EDF$，$\angle AFN=\angle CFD$，$\therefore \angle ABN+\angle BAF=\angle DEC+\angle FCE$，则$45^{\circ}+\angle FCE=\angle ABN+90^{\circ}$．$\because \angle BCE=\angle FCE+\angle BCA=\angle FCE+45^{\circ}$，$\therefore \angle ABN+90^{\circ}=\angle BCE$，且$\angle BCE=\angle ACD+\angle ECD+\angle BCA=\angle ACD+90^{\circ}$，$\therefore \angle ABN=\angle ACD$，$\therefore \triangle ABN\cong \triangle ACD(SAS)$，$\therefore AD=AN$，$\angle CAD=\angle BAN$，$\therefore \angle DAN=\angle CAD+\angle CAN=\angle CAN+\angle BAN=\angle BAC=90^{\circ}$，$\therefore \triangle ADN$为等腰直角三角形．在等腰直角三角形$ADN$中，$M$为$DN$的中点，$\therefore AM\perp DM$，$AM=DM$．