已知平行四边形ABCD,点E是AB的中点,在直线AD上截取AF=2FD,EF交AC于G,求 AG/AC?
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![已知平行四边形ABCD,点E是AB的中点,在直线AD上截取AF=2FD,EF交AC于G,求 AG/AC?](/uploads/image/z/3977847-63-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%B9%B3%E8%A1%8C%E5%9B%9B%E8%BE%B9%E5%BD%A2ABCD%2C%E7%82%B9E%E6%98%AFAB%E7%9A%84%E4%B8%AD%E7%82%B9%2C%E5%9C%A8%E7%9B%B4%E7%BA%BFAD%E4%B8%8A%E6%88%AA%E5%8F%96AF%EF%BC%9D2FD%2CEF%E4%BA%A4AC%E4%BA%8EG%2C%E6%B1%82+AG%EF%BC%8FAC%3F)
已知平行四边形ABCD,点E是AB的中点,在直线AD上截取AF=2FD,EF交AC于G,求 AG/AC?
已知平行四边形ABCD,点E是AB的中点,在直线AD上截取AF=2FD,EF交AC于G,求 AG/AC?
已知平行四边形ABCD,点E是AB的中点,在直线AD上截取AF=2FD,EF交AC于G,求 AG/AC?
2/7?
过E作AD的平行线交AC与H 交cd与M
AB//CD AD//EM 根据平行四边形的性质
所以△AEH与△CGM全等
EH=HM=1/2 EM =1/2 AD
AH= HC =1/2 AC
AF//EH
△AFG与△EGH相似
AG/GH=AF/EH
AF=2FD =2/3 AD
AG/GH =(2/3AD )/(1/2AD)=4/3
所以AG/AH=4/7
s所以AG/AC=2/7
(1)点F在线段AD上时,设EF与CD的延长线交于H,
∵AB∥CD,
∴△EAF∽△HDF,
∴HD:AE=DF:AF=1:2,
即HD=12AE
∵AB∥CD,
∴△CHG∽△AEG,
∴AG:CG=AE:CH
∵AB=CD=2AE,
∴CH=CD+DH=2AE+12AE=52AE,
∴AG:CG=2:5,
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(1)点F在线段AD上时,设EF与CD的延长线交于H,
∵AB∥CD,
∴△EAF∽△HDF,
∴HD:AE=DF:AF=1:2,
即HD=12AE
∵AB∥CD,
∴△CHG∽△AEG,
∴AG:CG=AE:CH
∵AB=CD=2AE,
∴CH=CD+DH=2AE+12AE=52AE,
∴AG:CG=2:5,
∴AG:(AG+CG)=2:(2+5),
即AG:AC=2:7;
(2)点F在线段AD的延长线上时,设EF与CD交于H,
∵AB∥CD,
∴△EAF∽△HDF,
∴HD:AE=DF:AF=1:2,
即HD=12AE,
∵AB∥CD,
∴△CHG∽△AEG,
∴AG:CG=AE:CH
∵AB=CD=2AE,
∴CH=CD-DH=2AE-12AE=32AE,
∴AG:CG=2:3,
∴AG:(AG+CG)=2:(2+3),
即AG:AC=2:5.
答案:填2/5或2/7.
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