《2025年通城学典非常课课通九年级数学上册人教版江苏专版》

3. 旋转类图形变化迁移
典例 12 通过类比联想、引申拓展研究典型题目，可达到解一题知一类的目的. 下面是一个案例，请补充完整.
(1)原题：如图①，点$E$，$F$分别在正方形$ABCD$的边$BC$，$CD$上，$\angle EAF = 45^{\circ}$，连接$EF$，则$EF = BE + DF$，请给出证明过程.
(2)类比联想：如图②，在四边形$ABCD$中，$AB = AD$，$\angle BAD = 90^{\circ}$，点$E$，$F$分别在边$BC$，$CD$上，$\angle EAF = 45^{\circ}$. 如果$\angle B$，$\angle D$都不是直角，那么当$\angle B$与$\angle D$满足等量关系时，仍有$EF = BE + DF$，并给出证明过程.
(3)引申拓展：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$BD$，$DE$，$EC$满足的等量关系，并写出推理过程.

答案：
解析：
(1)将$\triangle ABE$绕点$A$按逆时针方向旋转$90^{\circ}$，得到$\triangle ADG$，先证点$F$，$D$，$G$共线，再证$\triangle AEF\cong\triangle AGF$，得$EF = GF$，即可证得结论正确.
(2)将$\triangle ABE$绕点$A$按逆时针方向旋转$90^{\circ}$，得到$\triangle ADG$，先证点$F$，$D$，$G$共线，再证$\triangle AFE\cong\triangle AFG$，得$EF = GF$，即可证得结论正确.
(3)将$\triangle AEC$绕点$A$按顺时针方向旋转$90^{\circ}$，得到$\triangle AE'B$，连接$DE'$，证$\angle E'BD = 90^{\circ}$，由勾股定理得到$DE'^{2} = BD^{2}+BE'^{2}$，再证$\triangle AE'D\cong\triangle AED$，从而转化线段，得到$BD$，$DE$，$EC$的关系.
解：
(1)$\because AB = AD$，$\therefore AB$与$AD$能重合. 如图④，将$\triangle ABE$绕点$A$按逆时针方向旋转$90^{\circ}$，得到$\triangle ADG$. $\therefore\triangle ABE\cong\triangle ADG$. $\therefore AE = AG$，$BE = DG$，$\angle1 = \angle2$，$\angle B = \angle ADG = 90^{\circ}$. $\because\angle ADC+\angle ADG = 180^{\circ}$，$\therefore$点$F$，$D$，$G$共线. $\because\angle EAF = 45^{\circ}$，$\therefore\angle1+\angle3 = 45^{\circ}$. $\therefore\angle2+\angle3 = 45^{\circ}$. $\therefore\angle EAF = \angle GAF$. 在$\triangle AEF$和$\triangle AGF$中，$\begin{cases}AE = AG\\\angle EAF = \angle GAF\\AF = AF\end{cases}$，$\therefore\triangle AEF\cong\triangle AGF$. $\therefore EF = GF$. $\therefore EF = DG + DF = BE + DF$.
(2)$\angle B+\angle D = 180^{\circ}$. $\because AB = AD$，$\therefore AB$与$AD$能重合. 如图⑤，将$\triangle ABE$绕点$A$按逆时针方向旋转$90^{\circ}$，得到$\triangle ADG$. $\therefore\triangle ABE\cong\triangle ADG$. $\therefore AE = AG$，$BE = DG$，$\angle B = \angle5$，$\angle1 = \angle3$. $\because\angle B+\angle4 = 180^{\circ}$，$\therefore\angle5+\angle4 = 180^{\circ}$. $\therefore$点$F$，$D$，$G$共线. $\because\angle EAF = 45^{\circ}$，$\therefore\angle1+\angle2 = 45^{\circ}$. $\therefore\angle2+\angle3 = 45^{\circ}$. $\therefore\angle EAF = \angle GAF$. $\because AE = AG$，$AF = AF$，$\therefore\triangle AFE\cong\triangle AFG$. $\therefore EF = GF$. $\therefore EF = DG + DF = BE + DF$.
(3)$DE^{2} = BD^{2}+EC^{2}$. 如图⑥，将$\triangle AEC$绕点$A$按顺时针方向旋转$90^{\circ}$，得到$\triangle AE'B$，连接$DE'$. $\therefore\triangle AE'B\cong\triangle AEC$. $\therefore BE' = CE$，$AE' = AE$，$\angle1 = \angle C$，$\angle3 = \angle4$. $\because\angle BAC = 90^{\circ}$，$AB = AC$，$\therefore\angle2 = \angle C = 45^{\circ}$. $\therefore\angle2+\angle1 = 90^{\circ}$，即$\angle E'BD = 90^{\circ}$. $\therefore DE'^{2} = BD^{2}+BE'^{2}$. 又$\because\angle7 = 45^{\circ}$，$\therefore\angle8+\angle4 = 45^{\circ}$. $\therefore\angle3+\angle8 = 45^{\circ}$，即$\angle E'AD = 45^{\circ}$. 在$\triangle AE'D$和$\triangle AED$中，$\because AE' = AE$，$\angle E'AD = \angle EAD$，$AD = AD$，$\therefore\triangle AE'D\cong\triangle AED$. $\therefore DE' = DE$. $\therefore DE^{2} = BD^{2}+EC^{2}$.