A straight line is drawn through the point F of the bisector of the angle BAC, parallel to the straight line AC

A straight line is drawn through the point F of the bisector of the angle BAC, parallel to the straight line AC and intersecting the ray AB at point P. Calculate the degree measures of the angles of the triangle APF, if the angle FAC = 20 degrees.

Since AF is the bisector of the angle BAC, then the angle BAF = FAC = 20.

Since, by condition, the line PF is parallel to AC, the angle AFP is equal to the angle FAC as criss-crossing angles at the intersection of the secant AF of parallel straight lines AC and PF.

Then in the triangle APF the angle APF = (180 – PAF – AFP) = 180 – 20 – 20 = 140.

Answer: The angles of the triangle APF are equal to 20, 20, 140.