AB = AC，∠BAC= 120，

∴∠B=∠C= =30

∵ The middle vertical line EF of AC intersects AC at point E and BC at point F,

∴CF=AF (the distance between the point on the middle vertical line of the line segment and the two endpoints of the line segment is equal),

∴∠fac =∞∠c = 30 (equiangular), (2 points)

∴∠BAF =∠BAC-∠fac = 120-30 = 90，( 1)

In Rt△ABF, ∠ b = 30,

∴BF=2AF (in a right triangle, the right-angled side facing an angle of 30 is equal to half of the hypotenuse), (1 min)

∴BF=2CF (equivalent substitution).