In trapezoid ABCD AB = CD, the angle BDA = 49 degrees and the angle BDC = 13 degrees. Find the angle ABD.

Given:

isosceles trapezoid ABCD,

AB = CD,

angle BDA = 49 degrees,

angle BDC = 13 degrees.

Find the degree measure of the angle ABD -?

Decision:

1. Consider angle D. Angle D = angle CDB + angle BDA = 13 + 49 = 62 degrees.

2. Consider an isosceles trapezoid ABCD. Its angles at the base are equal to each other, that is, the angle A = angle D = 62 degrees.

3. Consider the triangle ABD. We know that the sum of the degree measures of the angles of a triangle is 180 degrees, that is

angle BAD + angle ADB + angle ABD = 180;

62 + 49 + angle ABD = 180;

111 + angle ABD = 180;

angle ABD = 180 – 111;

angle ABD = 69 degrees.

Answer: 69 degrees.