《2025年名校课堂八年级数学上册人教版》

6. （教材P43习题T1变式）（2024·西藏）如图，$C$是线段$AB$的中点，$AD = BE$，$\angle A = \angle B$。求证：$\angle D = \angle E$。

答案： 6.证明：$\because C$是线段$AB$的中点，$\therefore AC = BC$．在$\triangle DAC$和$\triangle EBC$中，$\begin{cases}AD = BE, \\\angle A = \angle B, \\AC = BC,\end{cases}$$\therefore \triangle DAC \cong \triangle EBC(SAS)$．$\therefore \angle D = \angle E$．

7. 新考向 开放性问题 如图，在$\triangle ABC$与$\triangle DEF$中，$\angle B = \angle E = 90^{\circ}$，点$B$，$F$，$C$，$E$在一条直线上，$AB = DE$。要使$\triangle ABC\cong\triangle DEF$，给出下列三个条件：①$AC = DF$；②$\angle A = \angle D$；③$BF = CE$。请从以上三个条件中选择一个合适的条件并证明$\triangle ABC\cong\triangle DEF$。（图形中不得添加任何字母）

答案： 7.解：选择①，证明如下：$\because \angle B = \angle E = 90^{\circ}$，$\therefore \triangle ABC$与$\triangle DEF$均为直角三角形．在$Rt\triangle ABC$和$Rt\triangle DEF$中，$\begin{cases}AC = DF, \\AB = DE,\end{cases}$ $\therefore Rt\triangle ABC \cong$$Rt\triangle DEF(HL)$．选择②，证明如下：在$\triangle ABC$和$\triangle DEF$中，$\begin{cases}\angle A = \angle D, \\AB = DE, \\\therefore \triangle ABC \cong \triangle DEF(ASA).\\\angle B = \angle E,\end{cases}$选择③，证明如下：$\because BF = CE$，$\therefore BC = EF$．在$\triangle ABC$和$\triangle DEF$中，$\begin{cases}AB = DE, \\\angle B = \angle E, \\BC = EF,\end{cases}$ $\therefore \triangle ABC \cong \triangle DEF$$(SAS)$．

8. 如图，$\angle C = \angle E$，$AC = AE$，点$D$在边$BC$上，$\angle 1 = \angle 2$，$AC$和$DE$相交于点$O$。求证：$\triangle ABC\cong\triangle ADE$。

答案： 8.证明：$\because \angle ADC = \angle ADE + \angle 2 = \angle 1 + \angle B$，且$\angle 1 = \angle 2$，$\therefore \angle ADE =$$\angle B$．在$\triangle ABC$和$\triangle ADE$中，$\begin{cases}\angle C = \angle E, \\\angle B = \angle ADE, \\AC = AE,\end{cases}$ $\therefore \triangle ABC \cong \triangle ADE$$(AAS)$．

9. 如图，$CB$为$\angle ACE$的平分线，$F$是线段$CB$上一点，$CA = CF$，$\angle B = \angle E$，延长$EF$与线段$AC$相交于点$D$。
（1）求证：$AB = FE$。
（2）若$ED\perp AC$，$AB// CE$，求$\angle A$的度数。

答案： 9.解：
(1)证明：$\because CB$为$\angle ACE$的平分线，$\therefore \angle ACB = \angle FCE$．在$\triangle ABC$和$\triangle FEC$中，$\begin{cases}\angle B = \angle E, \\\angle ACB = \angle FCE, \\CA = CF,\end{cases}$ $\therefore \triangle ABC \cong \triangle FEC(AAS)$．$\therefore AB = FE$．
(2)$\because AB // CE$，$\therefore \angle B = \angle FCE$．$\therefore \angle E = \angle B = \angle FCE = \angle ACB$．$\because ED$$\perp AC$，即$\angle CDE = 90^{\circ}$，$\therefore \angle E + \angle FCE + \angle ACB = 90^{\circ}$，即$3\angle ACB =$$90^{\circ}$．$\therefore \angle ACB = 30^{\circ}$．$\therefore \angle B = 30^{\circ}$．$\therefore \angle A = 180^{\circ} - \angle B - \angle ACB = 180^{\circ} -$