In parallelogram ABCD, the bisector of angle A meets side BC at point M. Prove that triangle ABM is isosceles.

Since AM is the bisector of the angle BAD, the angle BAM = DAM.

Angle BAM and DAM cross-lying angles at the intersection of parallel straight lines BC and AD secant AM, then the angle BMA = DAM.

Then the angle BAM = BMA = DAM. In triangle ABM, the angles at the base of AM are equal, therefore, triangle ABM is isosceles, AB = BM, which was required to be proved.