Segment AD – bisector of triangle ABC. A straight line is drawn through point D, parallel to side AB

Segment AD – bisector of triangle ABC. A straight line is drawn through point D, parallel to side AB and intersecting the side of AC at point H. Find the angles of the triangle ADH if the angle BAC is 72 degrees.

Since AD is the bisector of the angle BAC, then the angle BAD = CAD = ABC / 2 = 72/2 = 36.

By condition, DH is parallel to AB, then AH is a secant intersecting two parallel straight lines AB and DH, and then the angle BAD = ADH as criss-crossing angles, and then the angle DAН = ADH = 36.

Then in the triangle ADH the angles at the base of ADH are equal, and the triangle is isosceles.

Angle АНD = (180 – АDН – DАН) = (180 – 36 – 36) = 108.

Answer: The angles of the ADH triangle are 36, 36, 108.