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).