画一条直线将一个任意四边形分成面积相同的两部分

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2024-12-17 07:26:58
推荐回答(4个)
回答1:

任意四边形ABCD,取AC中点O,过O做BD平行线,交BC,DC于E,F,连接BF

BF就是所求直线

证明:S△ADO=S△DOC,S△ABO=S△BOC

S△ADO+S△ABO=S△DOC+S△BOC=1/2*S四边形ABCD=S四边形DABO

EF‖BD

S△DOB,S△DFB同底,同高

S△DBF=S△DOB

1/2*S四边形ABCD=S四边形DABO

=S△DOB+S△DAB=S△DBF+S△DAB

=S四边形DABF

BF把ABCD分成两个面积相同的部分

凹四边形

凹四边形四个顶点在同一平面内,对边不相交且作出一边所在直线,其余各边有些在其异侧。

依次连接四边形各边中点所得的四边形称为中点四边形。不管原四边形的形状怎样改变,中点四边形的形状始终是平行四边形。中点四边形的形状取决于原四边形的对角线。若原四边形的对角线垂直,则中点四边形为矩形;若原四边形的对角线相等,则中点四边形为菱形;若原四边形的对角线既垂直又相等,则中点四边形为正方形。

回答2:

任意四边形ABCD,取AC中点O,过O做BD平行线,交BC,DC于E,F,连接BF
BF就是所求直线
证明:S△ADO=S△DOC,S△ABO=S△BOC
S△ADO+S△ABO=S△DOC+S△BOC=1/2*S四边形ABCD=S四边形DABO
EF‖BD
S△DOB,S△DFB同底,同高
S△DBF=S△DOB
1/2*S四边形ABCD=S四边形DABO
=S△DOB+S△DAB=S△DBF+S△DAB
=S四边形DABF
BF把ABCD分成两个面积相同的部分

回答3:

将四边形沿对角线分成两个三角形,求出两三角形的重心,连接两个重心就可以了。

回答4:

只要过重心的直线都满足

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