As shown in the figure, ab = AC, ∠ BAC = 120 ° in △ ABC, the vertical bisector EF of AC intersects AC at point E and BC at point F. verification: BF = 2cf

It is proved that: the vertical bisector EF connecting AF, (1 point) ∵ AB = AC, ∠ BAC = 120 °, ∵ B = ∠ C = 180 − 120 ° 2 = 30 °, (1 point) ∵ AC intersects AC at point E, BC at point F, ∵ CF = AF (the distance from the point on the vertical bisector to the two ends of the line segment is equal), ∵ fac = ∠ C = 30 ° (equilateral and equal angle), (2 points) ∵ BAF = ∠ BAC - ∠ fac = 120 ° - 30 ° = 90 ° and (1 point) at RT △ In ABF, ∠ B = 30 ° BF = 2AF (in a right triangle, the right side opposite the angle of 30 ° is equal to half of the hypotenuse), (1 point) BF = 2cf (equivalent substitution)

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