题霸题霸学习平台
← 返回公开题库
八年级数学solution一般
题目
如图,DD,AA,EE三点都在一条直线上,且BDA=AEC=BAC\angle BDA=\angle AEC=\angle BAC,AB=ACAB=AC,求BDBD,CECE,DEDE之间的数量关系.
知识点:三角形综合题章节:未标注

答案与解析

答案

DE=BD+CEDE=BD+CE
理由如下:
BAE=D+ABD=BAC+CAE\because \angle BAE=\angle D+\angle ABD=\angle BAC+\angle CAE,且ADB=AEC=BAC\angle ADB=\angle AEC=\angle BAC
ABD=CAE\therefore \angle ABD=\angle CAE
ABD\triangle ABDCAE\triangle CAE中,
{ABD=CAEADB=CEAAB=AC\left\{\begin{array}{l}{∠ABD=∠CAE}\\{∠ADB=∠CEA}\\{AB=AC}\end{array}\right.
ABD\therefore \triangle ABDCAE(AAS)\triangle CAE\left(AAS\right)
AD=CE\therefore AD=CEBD=AEBD=AE
DE=AD+AE\because DE=AD+AE
DE=CE+BD\therefore DE=CE+BD.

解析

DE=BD+CEDE=BD+CE
理由如下:
BAE=D+ABD=BAC+CAE\because \angle BAE=\angle D+\angle ABD=\angle BAC+\angle CAE,且ADB=AEC=BAC\angle ADB=\angle AEC=\angle BAC
ABD=CAE\therefore \angle ABD=\angle CAE
ABD\triangle ABDCAE\triangle CAE中,
{ABD=CAEADB=CEAAB=AC\left\{\begin{array}{l}{∠ABD=∠CAE}\\{∠ADB=∠CEA}\\{AB=AC}\end{array}\right.
ABD\therefore \triangle ABDCAE(AAS)\triangle CAE\left(AAS\right)
AD=CE\therefore AD=CEBD=AEBD=AE
DE=AD+AE\because DE=AD+AE
DE=CE+BD\therefore DE=CE+BD.

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 →