6. 在$△ABC$中，$∠BAC = 90^{\circ}$，$AB = AC$，点$D$，$E$在直线$BC$上。如图①，若$∠DAE = 45^{\circ}$，求证：$BD^{2}+CE^{2}=DE^{2}$。
【阅读理解】
要证明$BD^{2}+CE^{2}=DE^{2}$，可设法将$BD$，$CE$，$DE$转化为某直角三角形的三边，故过点$A$作$AF⊥AD$，且$AF = AD$，连接$CF$，$EF$。再通过证明$△ABD≌△ACF$，$△AED≌△AEF$，即可将$BD$，$CE$，$DE$转化到直角$△ECF$中解决问题。
【拓展应用】
如图②，若$∠DAE = 135^{\circ}$，其他条件不变，请探究：以线段$BE$，$CD$，$DE$的长度为三边长的三角形是何种三角形？并说明理由。
【阅读理解】
要证明$BD^{2}+CE^{2}=DE^{2}$，可设法将$BD$，$CE$，$DE$转化为某直角三角形的三边，故过点$A$作$AF⊥AD$，且$AF = AD$，连接$CF$，$EF$。再通过证明$△ABD≌△ACF$，$△AED≌△AEF$，即可将$BD$，$CE$，$DE$转化到直角$△ECF$中解决问题。
【拓展应用】
如图②，若$∠DAE = 135^{\circ}$，其他条件不变，请探究：以线段$BE$，$CD$，$DE$的长度为三边长的三角形是何种三角形？并说明理由。
答案：
【阅读理解】证明:如答图①,过点A作$AF\perp AD$,且$AF = AD$,连接CF,EF.
$\because \angle DAE = 45^{\circ},\angle DAF = 90^{\circ}$,
$\therefore \angle DAE = \angle EAF = 45^{\circ}$,
在$\triangle EAD$和$\triangle EAF$中,$\left\{\begin{array}{l}EA = EA,\\\angle EAD = \angle EAF,\\AD = AF,\end{array}\right.$
$\therefore \triangle EAD\cong \triangle EAF,\therefore DE = EF$.
$\because \angle BAD + \angle CAE = 45^{\circ},\angle CAE + \angle CAF = 45^{\circ}$,
$\therefore \angle BAD = \angle CAF$,
在$\triangle BAD$和$\triangle CAF$中,$\left\{\begin{array}{l}AB = AC,\\\angle BAD = \angle CAF,\\AD = AF,\end{array}\right.$
$\therefore \triangle BAD\cong \triangle CAF$,
$\therefore BD = CF,\angle B = \angle ACF = 45^{\circ}$,
$\because \angle ACB = 45^{\circ},\therefore \angle ECF = 90^{\circ}$,
$\therefore EF^{2} = EC^{2} + CF^{2}$,
$\therefore DE^{2} = BD^{2} + CE^{2}$.
【拓展应用】解:以线段BE,CD,DE的长度为三边长的三角形是直角三角形.
理由:如答图②,作$AF\perp AE$,使得$AF = AE$,连接DF,CF.
$\because \angle EAF = \angle BAC = 90^{\circ}$,
$\therefore \angle FAC = \angle EAB$,
在$\triangle FAC$和$\triangle EAB$中,$\left\{\begin{array}{l}AF = AE,\\\angle FAC = \angle EAB,\\AC = AB,\end{array}\right.$
$\therefore \triangle FAC\cong \triangle EAB$,
$\therefore BE = CF,\angle ACF = \angle EBA = 45^{\circ}$,
$\because \angle ACB = 45^{\circ},\therefore \angle FCB = 90^{\circ}$,
$\because \angle DAE = 135^{\circ},\angle EAF = 90^{\circ}$,
$\therefore \angle DAF = 360^{\circ} - 135^{\circ} - 90^{\circ} = 135^{\circ}$,
$\therefore \angle DAF = \angle DAE$.
$\because AD = AD,AF = AE$,
$\therefore \triangle DAF\cong \triangle DAE,\therefore DF = DE$.
在$Rt\triangle DCF$中,
$\because DF^{2} = DC^{2} + CF^{2}$,
$\therefore DE^{2} = DC^{2} + BE^{2}$,
$\therefore$以线段BE,CD,DE的长度为三边长的三角形是直角三角形.
【阅读理解】证明:如答图①,过点A作$AF\perp AD$,且$AF = AD$,连接CF,EF.
$\because \angle DAE = 45^{\circ},\angle DAF = 90^{\circ}$,
$\therefore \angle DAE = \angle EAF = 45^{\circ}$,
在$\triangle EAD$和$\triangle EAF$中,$\left\{\begin{array}{l}EA = EA,\\\angle EAD = \angle EAF,\\AD = AF,\end{array}\right.$
$\therefore \triangle EAD\cong \triangle EAF,\therefore DE = EF$.
$\because \angle BAD + \angle CAE = 45^{\circ},\angle CAE + \angle CAF = 45^{\circ}$,
$\therefore \angle BAD = \angle CAF$,
在$\triangle BAD$和$\triangle CAF$中,$\left\{\begin{array}{l}AB = AC,\\\angle BAD = \angle CAF,\\AD = AF,\end{array}\right.$
$\therefore \triangle BAD\cong \triangle CAF$,
$\therefore BD = CF,\angle B = \angle ACF = 45^{\circ}$,
$\because \angle ACB = 45^{\circ},\therefore \angle ECF = 90^{\circ}$,
$\therefore EF^{2} = EC^{2} + CF^{2}$,
$\therefore DE^{2} = BD^{2} + CE^{2}$.
【拓展应用】解:以线段BE,CD,DE的长度为三边长的三角形是直角三角形.
理由:如答图②,作$AF\perp AE$,使得$AF = AE$,连接DF,CF.
$\because \angle EAF = \angle BAC = 90^{\circ}$,
$\therefore \angle FAC = \angle EAB$,
在$\triangle FAC$和$\triangle EAB$中,$\left\{\begin{array}{l}AF = AE,\\\angle FAC = \angle EAB,\\AC = AB,\end{array}\right.$
$\therefore \triangle FAC\cong \triangle EAB$,
$\therefore BE = CF,\angle ACF = \angle EBA = 45^{\circ}$,
$\because \angle ACB = 45^{\circ},\therefore \angle FCB = 90^{\circ}$,
$\because \angle DAE = 135^{\circ},\angle EAF = 90^{\circ}$,
$\therefore \angle DAF = 360^{\circ} - 135^{\circ} - 90^{\circ} = 135^{\circ}$,
$\therefore \angle DAF = \angle DAE$.
$\because AD = AD,AF = AE$,
$\therefore \triangle DAF\cong \triangle DAE,\therefore DF = DE$.
在$Rt\triangle DCF$中,
$\because DF^{2} = DC^{2} + CF^{2}$,
$\therefore DE^{2} = DC^{2} + BE^{2}$,
$\therefore$以线段BE,CD,DE的长度为三边长的三角形是直角三角形.