在平行四边形ABCD中,过点A的直线交对角线于点E,分别交射线BC,CD于点F,G,若EF=3,GF=2,则AE等于多少
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∵ABCD是平行四边形, ∴AD∥CF, ∴CG/DG=GF/AG,

∴(DC-DG)/DG=2/(AE+EG)=2/[AE+(EF-GF)]=2/[AE+(3-2)],

∴DC/DG-1=2/(AE+1), ∴DC/DG=1+2/(AE+1)=(AE+3)/(AE+1).······①

∵ABCD是平行四边形, ∴DG∥AB、DC=AB, ∴△DGE∽△BAE,

∴DG/AB=EG/AE, ∴DG/DC=(EF-GF)/AE=(3-2)/AE=1/AE.······②

①×②,得:1=(AE+3)/[AE(AE+1)], ∴AE(AE+1)=AE+3, ∴AE^2=3,

∴AE=√3.