# In an isosceles triangle ABC with base AB, the angle A = 60 (degrees) Prove that the bisector BN

**In an isosceles triangle ABC with base AB, the angle A = 60 (degrees) Prove that the bisector BN of the angle CBD adjacent to the angle B of the triangle is parallel to AC.**

Triangle ABC is isosceles by condition, which means that

∠ B = ∠ A = 60 ° (angles at the base of an isosceles triangle).

Find the degree measure of the CBD angle adjacent to the corners B.

∠ CBD = 180 ° – ∠ B = 120 °.

∠ NBC = ∠ CBD / 2 = 60 ° (BN is the bisector of the CBD angle).

∠ ABN = ∠ NBC + ∠ B = 120 °.

We got that the angle ABN is 120 °, the angle A is 60 °, their sum is 180 °.

These two angles are one-sided with straight lines AC, BN and secant AB, their sum 180 ° is one of the signs of parallelism of straight lines.

Straight lines BN and АС are parallel, as required.