In an isosceles triangle ABC with base BC, the bisector of the outer angle A is drawn.

In an isosceles triangle ABC with base BC, the bisector of the outer angle A is drawn. Prove that this bisector is parallel to BC.

Since the triangle ABC is isosceles, the angles at the base of the AC are equal, the angle BAC = BCA.

Let the angle BAC = BCA = X0.

The outer angle of a triangle is equal to the sum of its two inner angles that are not adjacent to it.

Angle СBD = (BAC + BCA) = 2 * X.

BE is the bisector of the outer angle, then the angle EBD = EBC = X0.

The angle EBC and ACB are equal to X0 and there are cross-lying angles at the intersection of the straight lines AC and BE secant BC, then the lines BE and AC are parallel, which was required to prove.