问题: △ABC中,中线AD,BE交于点G,FG‖AC,求DF/BD,DF/BC,GF/EC.
解答:
解:∵AD、BE是△ABC的中线
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==>G是△ABC的重心
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==>∵FG//AC==>△DGF∽△DAC
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==>DF/DC=DG/DA=1/3==>DC=BD==>DF/DC=DF/DB=1/3
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==>设DF=x,DB=3x==>BC=2BD=6x==>DF/BC=1/6
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==>∵△DGF∽△DAC,G是△ABC的重心
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==>GF/AC=DG/DA=1/3
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==>设GF=y,AC=3y==>EC=[1/2]AC=3y/2
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==>GF/EC=2/3
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