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八年级数学解答题一般
题目
如图,在ABC\triangle ABC中,DDEEBCBC上两点,且AB=BDAB=BD,CA=CECA=CE,BFBFCGCG分别平分ABC\angle ABCACB\angle ACB,BFBFCGCG交于OO点,求证:点OOADE\triangle ADE的三个顶点的距离相等.
知识点:三角形章节:第13章 三角形 / 13.1 三角形的概念

答案与解析

答案

证明:如图,连接OAOAODODOEOE,设ADADBFBF相交于点MMAEAECGCG相交于点NN
AB=BD\because AB=BD
ABD\therefore \triangle ABD为等腰三角形,
BF\because BF平分ABC\angle ABC
AM=DM\therefore AM=DMBFADBF\bot AD
BF\therefore BFADAD的垂直平分线,
OA=OD\therefore OA=OD
同理,CA=CE\because CA=CECGCG平分ACB\angle ACB
AN=EN\therefore AN=ENCNAECN\bot AE
OA=OE\therefore OA=OE
OA=OD=OE\therefore OA=OD=OE,即点OOADE\triangle ADE的三个顶点的距离相等.

解析

证明:如图,连接OAOAODODOEOE,设ADADBFBF相交于点MMAEAECGCG相交于点NN
AB=BD\because AB=BD
ABD\therefore \triangle ABD为等腰三角形,
BF\because BF平分ABC\angle ABC
AM=DM\therefore AM=DMBFADBF\bot AD
BF\therefore BFADAD的垂直平分线,
OA=OD\therefore OA=OD
同理,CA=CE\because CA=CECGCG平分ACB\angle ACB
AN=EN\therefore AN=ENCNAECN\bot AE
OA=OE\therefore OA=OE
OA=OD=OE\therefore OA=OD=OE,即点OOADE\triangle ADE的三个顶点的距离相等.

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