问题补充:
如图,AB是⊙O的直径,弦CA、BD的延长线相交于点E,EF垂直BA的延长线于点F.若∠EBA=30,EF=根号3,EA=2AC,求AF的长
答案:
连接BC设:圆O的半径=r,AF=x.
∵△AEF∽△ABC
∴EA/AB=AF/AC
2AC/2r=x/AC
AC^2=rx -------------(1)
∵EA^2=EF^2+AF^2
∴4 AC^2=3+x^2 ---------(2)
(1)代入(2)
4 rx=3+x^2
x^2-4 rx+3=0
x=2r±√(4 r^2-3) -------(3)
∵tg30°=EF/FB=√3/3
∴FB=EF*(3/√3)=3
∵AF+AB=FB=3
∴x+2r=3 ---------------(4)
(3)代入(4)
2r±√(4 r^2-3)+2r=3
[±√(4 r^2-3)]^2=(3-4r)^2
4 r^2-3=9-24r+16r^2
r^2-2r+1=0
(r-1)^2=0
r=1即:AF=FB-AB=3-2r=1
答:AF长度为1.