Angle BCD is external to triangle ABC, CE is the bisector of angle BCD, with CE parallel to AB

Angle BCD is external to triangle ABC, CE is the bisector of angle BCD, with CE parallel to AB. Prove that a triangle is ABC-isosceles.

Let the value of the angle ВСD = Х0.

Since CE is the bisector of the angle BCD, then the angle BCE = DCE = (X / 2) 0.

The angle ACB is adjacent to the angle BCD, the sum of which is 180, then the angle BCA = (180 – X) 0.

Since, according to the condition, AB is parallel to CE, then the angle BCE = ABC = (X / 2) as criss-crossing angles at the intersection of parallel lines AB and CE secant BC.

Angle BAC = DCE = (X / 2) 0, as the corresponding angles at the intersection of parallel lines AB and CE secant AC.

In the triangle ABC, the angle BAC = ABC, and therefore the triangle is isosceles with the base AB, which was required to be proved.