The bisector of angle A of triangle abc intersects side bc at point K. On side AB, point N is marked

The bisector of angle A of triangle abc intersects side bc at point K. On side AB, point N is marked, so that AN = NK. Find the angles of the triangle ANK if the angle ABC = 40, and the difference between the angles BAC and ACB is 20.

Let the value of the angle BAC = X0, then the value of the angle ACB = (X – 20) 0.

The sum of the interior angles of the triangle is 180, then (40 + X + X – 20) = 180.

2 * X = 160.

X = 160/2 = 80.

Angle BAC = 80, angle ACB = 80 – 20 = 60.

Since AK is a bisector, the angle BAK = BAC / 2 = 80/2 = 40.

The triangle ANK is isosceles, since, by condition, AN = NK, then the angle AKN = 40, and the angle ANK = (180 – 40 – 40) = 100.

Answer: The angles of triangle ANK are 40, 40, 100.