# In isosceles trapezoid ABCD, the bisector of angle B intersects the larger base AD at point

In isosceles trapezoid ABCD, the bisector of angle B intersects the larger base AD at point M. The BMD angle is 128 degrees. Find the angle D.

The angle AMB is adjacent to the angle BMD, the sum of which is 180, then the angle AMB = 180 – BMD 180 – 128 = 52.

Angle MBS = AMB as criss-crossing angles at the intersection of parallel lines AD and BC secant MB.

MBS = AMB = 52. Since the segment BM is the bisector of the angle ABC, then the angle ABM = MBC = 52, then the angle ABC = 2 * ABM = 2 * 52 = 104.

The sum of the angles of the trapezoid at the side is 180, then the angle BAD = 180 – ABC = 180 – 104 = 76.

In an isosceles trapezoid, the angles at the base are equal, then the angle ADC = BAD = 76.

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