Angle A of an isosceles triangle ABC with base BC is 36o; BF is the bisector of triangle ABC.

Angle A of an isosceles triangle ABC with base BC is 36o; BF is the bisector of triangle ABC. Find BC if AF = 2015.

Since △ ABC is isosceles, then ∠C = ∠B = (180 ° – ∠A) / 2 = (180 ° – 36 °) / 2 = 144 ° / 2 = 72 ° (by the theorem on the sum of the angles of a triangle).
Since BF is the bisector of ∠B, then ∠FBC = ∠FBA = ∠B / 2 = 72 ° / 2 = 36 °.
Consider △ BFA: ∠FBA = ∠FAB = 36 °, then △ BFA is an isosceles triangle with base AB ⇒ AF = BF = 2015.
Consider △ FBC: ∠FCB = 72 °, ∠FBC = 36 °, then ∠CFB = 180 ° – ∠FCB – ∠FBC = 180 ° – 72 ° – 36 ° = 72 °.
Since △ FBC ∠FCB = ∠CFB = 72 °, then this triangle is isosceles with base FC, so BF = BC = 2015.
Answer: BC = 2015.