高中竞赛题-几何

锐角三角形ABC中,AC>AB,D,E分别在AC,AB上,且CD=BE,BD,CE交于F,G是AC上的点,满足GF平行于∠BAC的角平分线.求证 CG=AB.证明 设BA的延长线与FG交于H,∠BAC的角平分线交AC于J.因为AJ平分∠BAC,FG∥AJ,所以∠HGA=∠GAJ=∠JAB=∠AHG.故AH=AG.直线HGF截△AEC,由梅内劳斯定理,有(AH/HE)*(EF/FC)*(CG/GA)=1即得:CG/HE=FC/EF (1)又直线BFD截△AEC,由梅内劳斯定理,有(AB/BE)*(EF/FC)*(CD/DA)=1由BE=CD得:AB/DA=FC/EF (2)由(1),(2)得:AB/DA=CG/HE<===>(AE+EB)/(AG+GD)=(CD+DG)/(HA+AE)<==>AE^2+AE(EB+HA)=GD^2+GD(AG+CD)结合AG+CD=EB+HA可得:0=AE^2-GD^2=AE(EB+HA)-GD*EB+HA)=(AE-GD)*(AE+GD+EB+HA)所以 AE=GE.故AB=AE+EB=GD+DC=CG. 收起