《2025年玩转全课程八年级数学全一册浙教版》

(1) 思路梳理.
$\because AB= CD$,$\therefore把\triangle ABE$绕点A逆时针旋转$90^{\circ }至\triangle ADG$,可使AB与AD重合.
$\because \angle ADC= \angle B= 90^{\circ }$,$\therefore \angle FDG= 180^{\circ }$,点F,D,G共线.其依据是平角的定义,易证$\triangle AFG\cong$$\triangle AFE$,得$EF= BE+DF$.
(2) 类比引申.
如图2,四边形ABCD中,$AB= AD$,$\angle BAD= 90^{\circ }$.点E,F分别在边BC,CD上,$\angle EAF= 45^{\circ }$.若$\angle B$,$\angle D$都不是直角,则当$\angle B与\angle D$满足等量关系$\angle B + \angle D = 180^{\circ}$时,仍有$EF= BE+DF$.
(3) 联想拓展.
如图3,在$\triangle ABC$中,$\angle BAC= 90^{\circ }$,$AB= AC$,点D,E均在边BC上,且$\angle DAE= 45^{\circ }$.猜想BD,DE,EC应满足的等量关系,并写出推理过程.
猜想：$DE^{2} = BD^{2} + EC^{2}$
证明：将$\triangle AEC$绕点$A$顺时针旋转$90^{\circ}$得到$\triangle ABF$，连接$FD$
$\because \triangle AEC \cong \triangle ABF$
$\therefore BF = EC$，$AF = AE$，$\angle ABF = \angle C$，$\angle BAF = \angle CAE$
$\because \angle BAC = 90^{\circ}$，$\angle DAE = 45^{\circ}$
$\therefore \angle BAD + \angle CAE = 45^{\circ}$
$\therefore \angle BAD + \angle BAF = 45^{\circ}$，即$\angle FAD = 45^{\circ} = \angle DAE$
在$\triangle AFD$和$\triangle AED$中
$\left\{\begin{array}{l}AF = AE\\\angle FAD = \angle EAD\\AD = AD\end{array}\right.$
$\therefore \triangle AFD \cong \triangle AED(SAS)$
$\therefore FD = DE$
$\because AB = AC$，$\angle BAC = 90^{\circ}$
$\therefore \angle ABC = \angle C = 45^{\circ}$
$\therefore \angle ABF = 45^{\circ}$
$\therefore \angle FBD = \angle ABC + \angle ABF = 90^{\circ}$
在$Rt\triangle FBD$中，$FD^{2} = BD^{2} + BF^{2}$
$\therefore DE^{2} = BD^{2} + EC^{2}$

答案：
(1) 平角的定义；$\triangle AFE$
(2) $\angle B + \angle D = 180^{\circ}$
(3) 猜想：$DE^{2} = BD^{2} + EC^{2}$
证明：将$\triangle AEC$绕点$A$顺时针旋转$90^{\circ}$得到$\triangle ABF$，连接$FD$
$\because \triangle AEC \cong \triangle ABF$
$\therefore BF = EC$，$AF = AE$，$\angle ABF = \angle C$，$\angle BAF = \angle CAE$
$\because \angle BAC = 90^{\circ}$，$\angle DAE = 45^{\circ}$
$\therefore \angle BAD + \angle CAE = 45^{\circ}$
$\therefore \angle BAD + \angle BAF = 45^{\circ}$，即$\angle FAD = 45^{\circ} = \angle DAE$
在$\triangle AFD$和$\triangle AED$中
$\left\{\begin{array}{l}AF = AE\\\angle FAD = \angle EAD\\AD = AD\end{array}\right.$
$\therefore \triangle AFD \cong \triangle AED(SAS)$
$\therefore FD = DE$
$\because AB = AC$，$\angle BAC = 90^{\circ}$
$\therefore \angle ABC = \angle C = 45^{\circ}$
$\therefore \angle ABF = 45^{\circ}$
$\therefore \angle FBD = \angle ABC + \angle ABF = 90^{\circ}$
在$Rt\triangle FBD$中，$FD^{2} = BD^{2} + BF^{2}$
$\therefore DE^{2} = BD^{2} + EC^{2}$