Find the degree measure of the angle ADC of an isosceles trapezoid ABCD if the diagonal AC makes

Find the degree measure of the angle ADC of an isosceles trapezoid ABCD if the diagonal AC makes angles equal to 25 ° and 55 ° with the base BC and the lateral side AB, respectively.

At the trapezoid, the bases are parallel, then the angle CAD = ACB = 25 as the cross-lying angles at the intersection of parallel lines AD and BC secant AC.

Then the angle BAD = BAC + DAC = 55 + 25 = 80.

Since the trapezoid is isosceles, then the angle ADC = BAD = 80.

Second way.

In triangle ABC we define the angle ABC. Angle ABC = (180 – BAC – BCA) = (180 – 55 – 25) = 100.

Since the trapezoid is isosceles, the angle BCD = ABC = 100.

The sum of the angles of the trapezoid at the side is 180, then the angle ADC = (180 – BCD) = (180 – 100) = 80.

Answer: The ADC angle is 80.