The bisector of the outer angle at the vertex A of the triangle ABC is parallel to its side BC.

The bisector of the outer angle at the vertex A of the triangle ABC is parallel to its side BC. Is it true that triangle ABC is isosceles?

1) Let the outer angle be <B1AB, and the bisector AA1 divides the angle in half, and each angle is equal to a = <B1AA1 = <A1AB and equal to the angle <ABC, since straight lines AA1 and BC are parallel.

2) Consider all the angles of the triangle ABC: <BAC = 180 – 2 * a; angle <ABC = a, on the basis of parallelism of straight lines АА1 and ВС.

3) Determine the third angle of the triangle <BCA = 180 – a – (180 – 2 * a) = 180 – 180 – a + 2 * a = a ..

4) That is, in a triangle, two equal angles turned out: <ABC = <BCA = a, and this is possible only in an isosceles triangle.