In trapezoid ABCD, it is known that AB = CD, angle BDA = 22 ° and angle BDC = 45 °. Find the angle ABD.

Determine the value of the angle ADC.

ADC = ADB + BDC = 22 + 45 = 67.

Since the sum of the angles at the lateral sides of the trapezoid is 180, the angle ВСD = 180 – 67 = 113.

Since, by condition, AB = CD, the trapezoid is isosceles and the angle BCD = CBA = 113.

The angle DBC is equal to the angle ADB, as intersecting angles, at the intersection of lines AD and BC parallel to the secant BD.

Angle DВС = ADB = 22.

Then the angle ABD = CBA – DBC = 113 – 22 = 91.

Answer: Angle ABD = 91.