题霸题霸学习平台
← 返回公开题库
八年级数学解答题一般
题目
如图,AB=CDAB=CD,AF=CEAF=CE,BE=DFBE=DF.求证:ABABCD.CD.
知识点:平行线的性质、全等三角形的判定与性质章节:第一章 三角形 / 1.3 探索三角形全等的条件

答案与解析

答案

证明:AF=CE\because AF=CE
AFEF=CEEF\therefore AF-EF=CE-EF
AE=CF\therefore AE=CF
ABE\triangle ABECDF\triangle CDF中,
{AB=CDAE=CFBE=DF\left\{\begin{array}{l}{AB=CD}{}\\{AE=CF}{}\\{BE=DF}{}\end{array}\right.
ABE\therefore \triangle ABECDF(SSS)\triangle CDF\left(SSS\right)
A=DCF\therefore \angle A=\angle DCF
AB\therefore ABCD.CD.

解析

证明:AF=CE\because AF=CE
AFEF=CEEF\therefore AF-EF=CE-EF
AE=CF\therefore AE=CF
ABE\triangle ABECDF\triangle CDF中,
{AB=CDAE=CFBE=DF\left\{\begin{array}{l}{AB=CD}{}\\{AE=CF}{}\\{BE=DF}{}\end{array}\right.
ABE\therefore \triangle ABECDF(SSS)\triangle CDF\left(SSS\right)
A=DCF\therefore \angle A=\angle DCF
AB\therefore ABCD.CD.

自由组卷

围绕这道题再组一份练习 →

完整试卷

浏览同年级试卷结构 →