The height BH of parallelogram ABCD cuts off an isosceles triangle from it. Find the degree measure of the angle ADC.

Given: ABCD – parallelogram; ВН – height; AНВ is an isosceles triangle.

Find: angle ADC.

Since ВН is the height, then the angle AНВ = 90 °, then the triangle AНВ is isosceles and rectangular, which means that the angles at the base of this triangle are equal and add up to 90 °, which means that the angle ВAН = angle ABН = 90 °: 2 = 45 °.

By the property of the parallelogram, angle A = angle C = 45 °, as well as angle B = angle D = (360 – 45 * 2) / 2 = (360 – 90) / 2 = 270/2 = 135 ° – by the property of opposite angles in parallelogram (in a parallelogram the sum of all angles is 360 °).

Answer: angle ADC = 135 °.