《2025年名校课堂八年级数学上册人教版》

1. 如图所示，$AB = AC$，$BD = CE$，$AD = AE$。求证：$\triangle ABE\cong\triangle ACD$。

答案： 1.证明：$\because BD = CE$，$\therefore BD + DE = CE + DE$．$\therefore BE = CD$．在$\triangle ABE$和$\triangle ACD$中，$\begin{cases}AE = AD, \\AB = AC, \\\therefore\triangle ABE \cong \triangle ACD(SSS).\\BE = CD,\end{cases}$

2. （2024·云南）如图，在$\triangle ABC$和$\triangle AED$中，$AB = AE$，$\angle BAE = \angle CAD$，$AC = AD$。求证：$\triangle ABC\cong\triangle AED$。

答案： 2.证明：$\because \angle BAE = \angle CAD$，$\therefore \angle BAE + \angle CAE = \angle CAD + \angle CAE$，即$\angle BAC = \angle EAD$．在$\triangle ABC$和$\triangle AED$中，$\begin{cases}AB = AE, \\\angle BAC = \angle EAD, \\AC = AD,\end{cases}$$\therefore \triangle ABC$$\cong \triangle AED(SAS)$．

3. 如图，点$C$，$E$，$F$，$B$在同一条直线上，$DF\perp BC$，$AE// DF$，$AB = CD$，$AE = DF$。求证：$CE = BF$。

答案： 3.证明：$\because DF \perp BC$，$\therefore \angle DFC = 90^{\circ}$．$\because AE // DF$，$\therefore \angle AEB = \angle DFC =90^{\circ}$．在$Rt\triangle AEB$和$Rt\triangle DFC$中，$\begin{cases}AB = DC, \\AE = DF,\end{cases}$ $\therefore Rt\triangle AEB \cong Rt\triangle DFC$$(HL)$．$\therefore BE = CF$．$\therefore BE - EF = CF - EF$，即$CE = BF$．

4. 新考向 开放性问题 如图，$\triangle ABC$的顶点$A$，$B$和$\triangle DEF$的顶点$D$，$E$在同一条直线上，且$\angle A = \angle EDF$，$\angle C = \angle F$，请再添加一个条件，使得$BC = EF$，并说明理由。

答案： 4.解：答案不唯一．例如添加的条件为$AC = DF$．理由：在$\triangle ABC$和$\triangle DEF$中，$\begin{cases}\angle A = \angle EDF, \\AC = DF, \\\angle C = \angle F,\end{cases}$ $\therefore \triangle ABC \cong \triangle DEF(ASA)$．$\therefore BC = EF$．

5. 如图，已知点$E$，$C$在线段$BF$上，$BE = CF$，$\angle B = \angle DEF$，$\angle BAC = \angle EDF$。
（1）求证：$\triangle ABC\cong\triangle DEF$。
（2）连接$AD$，若$\angle CAD = \angle F = 68^{\circ}$，$DE$平分$\angle ADF$，求$\angle DEF$的度数。

答案： 5.解：
(1)证明：$\because BE = CF$，$\therefore BE + EC = CF + EC$，即$BC = EF$．在$\triangle ABC$和$\triangle DEF$中，$\begin{cases}\angle BAC = \angle EDF, \\\angle B = \angle DEF, \\BC = EF,\end{cases}$ $\therefore \triangle ABC \cong \triangle DEF(AAS)$．
(2)由
(1)知，$\triangle ABC \cong \triangle DEF$，$\therefore \angle ACB = \angle F$．$AC // DF$．$\because \angle CAD = \angle F =68^{\circ}$，$\therefore \angle ADF = 180^{\circ} - \angle CAD = 112^{\circ}$．$\because DE$平分$\angle ADF$，$\therefore \angle ADE =$$\angle DEF = \frac{1}{2}\angle ADF = 56^{\circ}$．$\therefore \angle DEF = 180^{\circ} - \angle EDF - \angle F = 180^{\circ} - 56^{\circ}$$68^{\circ} = 56^{\circ}$．