Question

如图,在△ABC中,AB>AC,AD为∠BAC的平分线,AD的垂直平分线交BC的延长线于F,分别交AB、AC于E、G

Answer 1

(1)证明:连接DG,EF垂直平分AD,则AG=DG;DF=AF.(线段垂直平分线的性质)

∴∠FAD=∠FDA;∠GDA=∠GAD.则:∠FAG=∠FDG.

又∠GAD=∠EAD,则∠GDA=∠EAD,DG∥BA,∠B=∠FDG=∠FAG;

又∠AFC=∠BFA,故⊿AFC∽⊿BFA,则AF/BF=CF/AF,AF^2=BF*CF.

∴DF²=BF*CF.

⊿AFC∽⊿BFA,则:CF/AF=AC/AB=3:4,设CF=3X,则AF=4X.

AF^2=BF*CF(已证),即:(4X)^2=BF*3X,BF=(16/3)X.

CF:BF=(3X):[(16/3)X]=9:16.

⊿AFC∽⊿BFA,则AB/AC=BF/AF,即X=Y/AF,AF=Y/X.

又AF^2=BF*CF,即(Y/X)^2=Y*(Y-6),化简得:

Y=(6X²)/(X²-1).其中(X>1)