任意六边形ABCDEF的各边中点一次为GHIJKL相对的中点连线GJ HK IL两两相交于 具体看图
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/08/24 01:41:58
任意六边形ABCDEF的各边中点一次为GHIJKL相对的中点连线GJ HK IL两两相交于 具体看图
![](http://img.wesiedu.com/upload/d/84/d84af1317a9296efc4cccf5e93135dae.jpg)
![](http://img.wesiedu.com/upload/d/84/d84af1317a9296efc4cccf5e93135dae.jpg)
证明:S_GHJK=S_GHJ+S_GKJ=1/2(S_GBJ+S_GCJ+S_GFJ+S_GEJ)
S_GIJL=S_GIJ+S_GLJ=1/2(S_GAJ+S_GFJ+S_GCJ+S_GDJ)=1/2(S_GBJ+S_GFJ+S_GCJ+S_GEJ)
故S_GHJK=S_GIJL 同理有S_GHJK=S_GIJL=S_HIKL
故S_QHK=((GQ-GP)/GJ ) *S_GHJK=(GQ/GJ)*S_GIJL -(GP/GJ)*S_GHJK=S_GLI -S_GKH
S_GLI=1/2(S_LAI+S_LBI)=1/4(S_LAC+S_LAD+S_LBC+S_LBD)=1/8(S_FAC+S_FAD+S_ABC+S_FBC
+S_BDA+S_BDF)
S_GKH=1/2(S_KAH+S_KBH)=1/4(S_FAH+S_EAH+S_FBH+S_EBH)=1/8(S_FAB+S_FAC+S_EAB
+S_EAC+S_FBC+S_EBC)
故S_GLI -S_GKH=1/8(S_FAD+S_ABC+S_BDA -S_FAB -S_EAB -S_EBC)+1/8(S_BDF -S_EAC)
=1/8(S_FAED -S_FED+S_ABC+S_ABCD-S_BCD -S_FAB -S_EFAB+S_EFA -S_BCDE+S_CDE)
+1/8(S_BDF-S_EAC)
=1/8(S_FADE+S_ABCD -S_FED -S_DCB -S_BAF) -1/8(S_EFAB+S_BCDE -S_ABC -S_EFA -S_CDE)
+1/8(S_BDF-S_EAC)
=1/8S_BDF -1/8S_EAC+1/8(S_BDF-S_EAC)=1/4(S_BDF-S_EAC)
即4S_QHK=S_BDF-S_EAC 同理有4S_PLI=S_BDF-S_EAC,4S_RGJ=S_BDF-S_EAC
即4S_QHK=4S_PLI=4S_RGJ=S_BDF-S_EAC
即S_QHK+S_PLI+S_RGJ=3/4(S_BDF-S_EAC)
S_GIJL=S_GIJ+S_GLJ=1/2(S_GAJ+S_GFJ+S_GCJ+S_GDJ)=1/2(S_GBJ+S_GFJ+S_GCJ+S_GEJ)
故S_GHJK=S_GIJL 同理有S_GHJK=S_GIJL=S_HIKL
故S_QHK=((GQ-GP)/GJ ) *S_GHJK=(GQ/GJ)*S_GIJL -(GP/GJ)*S_GHJK=S_GLI -S_GKH
S_GLI=1/2(S_LAI+S_LBI)=1/4(S_LAC+S_LAD+S_LBC+S_LBD)=1/8(S_FAC+S_FAD+S_ABC+S_FBC
+S_BDA+S_BDF)
S_GKH=1/2(S_KAH+S_KBH)=1/4(S_FAH+S_EAH+S_FBH+S_EBH)=1/8(S_FAB+S_FAC+S_EAB
+S_EAC+S_FBC+S_EBC)
故S_GLI -S_GKH=1/8(S_FAD+S_ABC+S_BDA -S_FAB -S_EAB -S_EBC)+1/8(S_BDF -S_EAC)
=1/8(S_FAED -S_FED+S_ABC+S_ABCD-S_BCD -S_FAB -S_EFAB+S_EFA -S_BCDE+S_CDE)
+1/8(S_BDF-S_EAC)
=1/8(S_FADE+S_ABCD -S_FED -S_DCB -S_BAF) -1/8(S_EFAB+S_BCDE -S_ABC -S_EFA -S_CDE)
+1/8(S_BDF-S_EAC)
=1/8S_BDF -1/8S_EAC+1/8(S_BDF-S_EAC)=1/4(S_BDF-S_EAC)
即4S_QHK=S_BDF-S_EAC 同理有4S_PLI=S_BDF-S_EAC,4S_RGJ=S_BDF-S_EAC
即4S_QHK=4S_PLI=4S_RGJ=S_BDF-S_EAC
即S_QHK+S_PLI+S_RGJ=3/4(S_BDF-S_EAC)
如图,G是正六边形ABCDEF的边CD的中点,连接AG交CE于点M,则GM:MA=______.
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(100分)正六边形ABCDEF,G为CD中点,连接AG,CE,交于M,求GM:MA比值
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电学 如图所示,ABCDEF为正六边形六个顶点,P,Q,M分别为AB ED AF的中点
证明:任意四边形的各边中点连线所成的四边形是平行四边形?
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