In triangle ABC, sides AB and BC are equal, and the angle BAC = 60 degrees, CF is the bisector

In triangle ABC, sides AB and BC are equal, and the angle BAC = 60 degrees, CF is the bisector of the angle adjacent to the angle ACB. Prove that line AB is parallel to line CF.

Since, by condition, AB = BC, then triangle ABC is isosceles, and as one of the angles of an isosceles triangle is 60, then triangle ABC is equilateral, and all of its internal angles are 60.

Angle ACB = 60.

The angle DCA is adjacent to the angle ACB, then the angle DCA = 180 – 60 = 120.

СF is the bisector of the adjacent angle, then the angle DCF = ACF = DAC / 2 = 120/2 = 60.

Angle АСF = BAC = 60, and since these are cross-lying angles at the intersection of straight lines AB and CF of the secant AC, then AB is parallel to CF, which was required to be proved.