1. 如图，已知$\angle ACB=90^{\circ }$，$AC=BC$，$AD\perp CE$，$BE\perp CE$，垂足分别是$D$，$E$，若$AD=3$，$BE=1$，则$DE$的长是____。
答案：2 点拨：$\because BE\perp CE,AD\perp CE,\therefore \angle E=\angle ADC=90^{\circ },$
$\therefore \angle EBC+\angle BCE=90^{\circ }.\because \angle BCE+\angle ACD=90^{\circ },$
$\therefore \angle EBC=\angle DCA.$
在$\triangle CEB$和$\triangle ADC$中，$\left\{\begin{array}{l} \angle E=\angle ADC,\\ \angle EBC=\angle DCA,\\ BC=AC,\end{array}\right. $
$\therefore \triangle CEB\cong \triangle ADC(AAS),\therefore BE=DC=1,CE=AD=3.$
$\therefore DE=EC-CD=3-1=2.$
$\therefore \angle EBC+\angle BCE=90^{\circ }.\because \angle BCE+\angle ACD=90^{\circ },$
$\therefore \angle EBC=\angle DCA.$
在$\triangle CEB$和$\triangle ADC$中，$\left\{\begin{array}{l} \angle E=\angle ADC,\\ \angle EBC=\angle DCA,\\ BC=AC,\end{array}\right. $
$\therefore \triangle CEB\cong \triangle ADC(AAS),\therefore BE=DC=1,CE=AD=3.$
$\therefore DE=EC-CD=3-1=2.$
2. 在直线$m$上依次取互不重合的三个点$D$，$A$，$E$，在直线$m$上方有$AB=AC$，且满足$\angle BDA=\angle AEC=\angle BAC=\alpha$。
(1)如图①，当$\alpha =90^{\circ }$时，线段$DE$，$BD$，$CE$之间的数量关系是
(2)如图②，当$0^{\circ }＜\alpha ＜180^{\circ }$时，(1)中结论是否仍然成立？若成立，请给出证明；若不成立，请说明理由。
(3)拓展与应用：如图③，当$\alpha =120^{\circ }$时，点$F$为$\angle BAC$平分线上的一点，且$AB=AF$，分别连接$FB$，$FD$，$FE$，$FC$，试判断$\triangle DEF$的形状，并说明理由。
(1)如图①，当$\alpha =90^{\circ }$时，线段$DE$，$BD$，$CE$之间的数量关系是
$DE=BD+CE$
。(2)如图②，当$0^{\circ }＜\alpha ＜180^{\circ }$时，(1)中结论是否仍然成立？若成立，请给出证明；若不成立，请说明理由。
解：$DE=BD+CE$仍然成立.
证明：$\because \angle BDA=\angle BAC=\angle AEC=\alpha ,$
$\therefore \angle BAD+\angle EAC=\angle BAD+\angle DBA=180^{\circ }-\alpha ,$
$\therefore \angle DBA=\angle EAC,\because AB=AC,$
$\therefore \triangle DBA\cong \triangle EAC(AAS),\therefore BD=AE,AD=CE,$
$\therefore DE=AD+AE=BD+CE.$
证明：$\because \angle BDA=\angle BAC=\angle AEC=\alpha ,$
$\therefore \angle BAD+\angle EAC=\angle BAD+\angle DBA=180^{\circ }-\alpha ,$
$\therefore \angle DBA=\angle EAC,\because AB=AC,$
$\therefore \triangle DBA\cong \triangle EAC(AAS),\therefore BD=AE,AD=CE,$
$\therefore DE=AD+AE=BD+CE.$
(3)拓展与应用：如图③，当$\alpha =120^{\circ }$时，点$F$为$\angle BAC$平分线上的一点，且$AB=AF$，分别连接$FB$，$FD$，$FE$，$FC$，试判断$\triangle DEF$的形状，并说明理由。
解：$\triangle DEF$是等边三角形. 理由：
$\because \alpha =120^{\circ }$，AF平分$\angle BAC,\therefore \angle BAF=\angle CAF=60^{\circ }.$
$\because AB=AF=AC,\therefore \triangle ABF$和$\triangle ACF$都是等边三角形，
$\therefore FA=FC,\angle FCA=\angle FAB=\angle AFC=60^{\circ },$
同(2)可得，$\triangle BDA\cong \triangle AEC,$
$\therefore \angle BAD=\angle ACE,AD=CE,$
$\therefore \angle FAD=\angle FCE,\therefore \triangle FAD\cong \triangle FCE(SAS),$
$\therefore DF=EF,\angle DFA=\angle EFC,$
$\therefore \angle DFE=\angle DFA+\angle AFE=\angle EFC+\angle AFE=\angle AFC=60^{\circ },\therefore \triangle DEF$是等边三角形.
$\because \alpha =120^{\circ }$，AF平分$\angle BAC,\therefore \angle BAF=\angle CAF=60^{\circ }.$
$\because AB=AF=AC,\therefore \triangle ABF$和$\triangle ACF$都是等边三角形，
$\therefore FA=FC,\angle FCA=\angle FAB=\angle AFC=60^{\circ },$
同(2)可得，$\triangle BDA\cong \triangle AEC,$
$\therefore \angle BAD=\angle ACE,AD=CE,$
$\therefore \angle FAD=\angle FCE,\therefore \triangle FAD\cong \triangle FCE(SAS),$
$\therefore DF=EF,\angle DFA=\angle EFC,$
$\therefore \angle DFE=\angle DFA+\angle AFE=\angle EFC+\angle AFE=\angle AFC=60^{\circ },\therefore \triangle DEF$是等边三角形.
答案：(1)$DE=BD+CE$
(2)解：$DE=BD+CE$仍然成立.
证明：$\because \angle BDA=\angle BAC=\angle AEC=\alpha ,$
$\therefore \angle BAD+\angle EAC=\angle BAD+\angle DBA=180^{\circ }-\alpha ,$
$\therefore \angle DBA=\angle EAC,\because AB=AC,$
$\therefore \triangle DBA\cong \triangle EAC(AAS),\therefore BD=AE,AD=CE,$
$\therefore DE=AD+AE=BD+CE.$
(3)解：$\triangle DEF$是等边三角形. 理由：
$\because \alpha =120^{\circ }$，AF平分$\angle BAC,\therefore \angle BAF=\angle CAF=60^{\circ }.$
$\because AB=AF=AC,\therefore \triangle ABF$和$\triangle ACF$都是等边三角形，
$\therefore FA=FC,\angle FCA=\angle FAB=\angle AFC=60^{\circ },$
同(2)可得，$\triangle BDA\cong \triangle AEC,$
$\therefore \angle BAD=\angle ACE,AD=CE,$
$\therefore \angle FAD=\angle FCE,\therefore \triangle FAD\cong \triangle FCE(SAS),$
$\therefore DF=EF,\angle DFA=\angle EFC,$
$\therefore \angle DFE=\angle DFA+\angle AFE=\angle EFC+\angle AFE=\angle AFC=60^{\circ },\therefore \triangle DEF$是等边三角形.
(2)解：$DE=BD+CE$仍然成立.
证明：$\because \angle BDA=\angle BAC=\angle AEC=\alpha ,$
$\therefore \angle BAD+\angle EAC=\angle BAD+\angle DBA=180^{\circ }-\alpha ,$
$\therefore \angle DBA=\angle EAC,\because AB=AC,$
$\therefore \triangle DBA\cong \triangle EAC(AAS),\therefore BD=AE,AD=CE,$
$\therefore DE=AD+AE=BD+CE.$
(3)解：$\triangle DEF$是等边三角形. 理由：
$\because \alpha =120^{\circ }$，AF平分$\angle BAC,\therefore \angle BAF=\angle CAF=60^{\circ }.$
$\because AB=AF=AC,\therefore \triangle ABF$和$\triangle ACF$都是等边三角形，
$\therefore FA=FC,\angle FCA=\angle FAB=\angle AFC=60^{\circ },$
同(2)可得，$\triangle BDA\cong \triangle AEC,$
$\therefore \angle BAD=\angle ACE,AD=CE,$
$\therefore \angle FAD=\angle FCE,\therefore \triangle FAD\cong \triangle FCE(SAS),$
$\therefore DF=EF,\angle DFA=\angle EFC,$
$\therefore \angle DFE=\angle DFA+\angle AFE=\angle EFC+\angle AFE=\angle AFC=60^{\circ },\therefore \triangle DEF$是等边三角形.