如图1,在Rt△ABC中,∠BAC=90°,AD⊥BC于点D,点O是AC边上一点,连接BO交AD于F,OE⊥OB交BC边

(3)OF/OE=n
设AB=2x,则：AC=2nx,AO=OC=nx,BC=2√(1+n^2)x
OB=√(4+n^2)x
BD=AB*sinC=2x/√(1+n^2)
在△BOC中,设∠OBC=a,则cosa=(BC^2+OB^2-OC^2)/2BC*OB=(2+n^2)/√(1+n^2)(4+n^2)
在RT△BDF中,BF=BD/cosa=2x√(4+n^2)/(2+n^2)
△ABF∽△COE
∴OE/OC=BF/AB
OE=BF*OC/AB=nx√(4+n^2)/(2+n^2)
OF=OB-BF=n^2x√(4+n^2)/(2+n^2)
∴OF/OE=n