In an isosceles triangle ABC (AC-Base), the bisector of the angle A-ray AD is drawn. Find the outer angle

In an isosceles triangle ABC (AC-Base), the bisector of the angle A-ray AD is drawn. Find the outer angle of the triangle at the vertex C, if the angle BAD = 35.

Since, by condition, AD is the bisector of the angle ABC, then the angle ABC = 2 * BAD = 2 * 35 = 70.

The ABC triangle is isosceles, then its angles at the base of the AC are equal, the angle BAC = BAC = 70.

The outer angle BCE, at the vertex C, adjacent to the inner angle of the BCA, the sum of which is 180, then the angle BCE = (180 – BCA) = 180 – 70 = 110.

Answer: The outer angle at the vertex C is 110.