In the triangle ABC ∠A = 50 °, ∠B = 65 °. A straight line BK is drawn through the vertex B so

In the triangle ABC ∠A = 50 °, ∠B = 65 °. A straight line BK is drawn through the vertex B so that the beam BС is the bisector of the angle ABK. Prove that AC║BK

Since BK is the bisector of the angle ABK, the value of the angle ABK = 2 * ABC = 2 * 65 = 130.

Extend straight line AB to point E. Angle DBE and ABD are adjacent angles, the sum of which is 180, then the angle KBE = 180 – ABK = 180 – 130 = 50.

Then the angle CAB = KBE.

Since the straight lines AC and BK intersect the straight line AE at the same angle, the straight line AC is parallel to BD, which was required to be proved.