《2025年高效精练八年级数学上册苏科版》

8. (2025·怒江州模拟)如图,点 D,C 在线段 AF 上,$AD=CF,AB=DE,BC=EF$.
求证:$△ABC\cong △DEF$.

证明：$\because AD = CF,\therefore AD + DC = CF + DC,\therefore AC = DF$，在$\triangle ABC$和$\triangle DEF$中，$\left\{\begin{array}{l} AB = DE,\\ BC = EF,\\ AC = DF,\end{array}\right. \therefore \triangle ABC\cong \triangle DEF(SSS)$.

答案： 证明：$\because AD = CF,\therefore AD + DC = CF + DC,\therefore AC = DF$，在$\triangle ABC$和$\triangle DEF$中，$\left\{\begin{array}{l} AB = DE,\\ BC = EF,\\ AC = DF,\end{array}\right. \therefore \triangle ABC\cong \triangle DEF(SSS)$。

9. (2024 秋·安次区期末)已知:如图,$△ABC$是等边三角形,D 是 AC 上一点,$∠ABD=∠ACE,AE// BC$.求证:$△ABD\cong △ACE$.

证明:$\because \triangle ABC$是等边三角形，$\therefore AB = AC,\angle BAC = 60^{\circ},\angle ACB = 60^{\circ}$。$\because AE// BC,\therefore \angle CAE = \angle ACB = 60^{\circ}$，$\therefore \angle BAD = \angle CAE$，在$\triangle ABD$和$\triangle ACE$中，$\left\{\begin{array}{l} \angle BAD = \angle CAE,\\ AB = AC,\\ \angle ABD = \angle ACE,\end{array}\right. \therefore \triangle ABD\cong \triangle ACE(ASA)$.

答案： $\because \triangle ABC$是等边三角形，$\therefore AB = AC,\angle ACB = \angle BAC$。$\because AE// BC,\therefore \angle CAE = \angle ACB$，$\therefore \angle BAD = \angle CAE$，在$\triangle ABD$和$\triangle ACE$中，$\left\{\begin{array}{l} \angle BAD = \angle CAE,\\ AB = AC,\\ \angle ABD = \angle ACE,\end{array}\right. \therefore \triangle ABD\cong \triangle ACE(ASA)$。

10. (2024 秋·市中区期末)如图,D,E 分别是 AB,AC 上的点,$AD=AE$,请添加一个条件,使得$△ABE\cong △ACD$.这个条件可以为$AB = AC$(只填一个条件即可).

答案： $AB = AC$，或$\angle AEB = \angle ADC$，或$\angle B = \angle C$。

11. (2024 秋·绵阳期末)如图,在$△ABC$中,D,E 两点分别在 AB,BC 边上,且$BD=BE$,现增加一个条件,使得$△ABE\cong △CBD$一定成立,则该条件可以是下列中的______①②③.
①$CE=AD$;②$∠BAC=∠BCA$;③$∠BAE=∠BCD$;④$AE⊥CD$.

答案： ①②③

12. (2025·扬州模拟)如图,在四边形 ABCD 中,$AB// CD$,在 BD 上取两点 E,F,使$DF=BE$,连接 AE,CF.
(1) 若$AE// CF$,试说明$△ABE\cong △CDF$;
(2) 在(1)的条件下,连接 AF,CE,试判断 AF 与 CE 有怎样的数量关系,并说明理由.

答案：

(1) 证明：$\because AB// CD,\therefore \angle ABD = \angle CDF$。$\because AE// CF,\therefore \angle AEB = \angle CFD$。$\because BF = DE,\therefore BF + EF = DE + EF,\therefore BE = DF$，在$\triangle ABE$和$\triangle CDF$中，$\left\{\begin{array}{l} \angle ABE = \angle CDF,\\ BE = DF,\\ \angle AEB = \angle CFD,\end{array}\right. \therefore \triangle ABE\cong \triangle CDF(ASA)$；
(2) 解：$AF = CE$，理由如下：如图，$\because \triangle ABF\cong \triangle CDE,\therefore AB = CD,AE = CF$，在$\triangle ABF$和$\triangle CDE$中，$\left\{\begin{array}{l} AB = CD,\\ \angle ABD = \angle CDB,\\ BF = DE,\end{array}\right. \therefore \triangle ABF\cong \triangle CDE(SAS),\therefore AF = CE$。