In the trapezoid ABCD AB = CD, the angle BDA = 54 degrees and the angle BDC = 33 degrees, find the angle ABD.

Given:

isosceles ABCD

AB = CD,

angle ВDА = 54 degrees,

angle ВDC = 33 degrees.

Find the degree measure of the angle ABD -?

Solution:

1. Angle D = angle CDB + angle BDA;

angle D = 33 + 54;

angle D = 87 degrees.

2. Consider an isosceles trapezoid ABCD. It has an angle D = angle A = 87 degrees and angle B = angle C. We know that the sum of the degree measures of a parallelogram is 360 degrees. Then

angle B + angle C + angle A + angle D = 360;

angle B + angle C + 87 + 87 = 360;

angle B + angle C + 174 = 360;

angle B + angle C = 360 – 174;

angle B + angle C = 186;

angle B = angle C = 93 degrees.

3. Consider the triangle ABD.

Angle ABD = 180 – 87 – 54;

Angle ABD = 39 degrees.

Answer: 39 degrees.