In the trapezium ABCD, it is known that AB =CD, angle BDA = 22 degrees and angle BDC = 45 degrees. Find the angle of the AED.

Let a trapezoid ABCD be given, in which BC ll AD, lateral sides AB = CD, angle углаADB = 22º, angle ºBDC = 45º.
The beam DВ passes between the sides of ∠ADС, which means that ∠ADС = ∠ADВ + ∠ВDC or ∠ADС = 22º + 45º = 67º.
By the property of an isosceles trapezoid, its angles at the bases are equal, that is, ∠BAD = ∠ADC = 67º.
The angles ∠BAD and ∠ABS are internal one-sided angles with BC ll AD and secant AB, then ∠BAD + ∠ABS = 180º. Let’s find the value of the angle ∠ABS = 180º – 67º = 113º.
The angles ∠ADB = ∠DBC = 22º, as the internal criss-crossing angles at ВС ll AD and secant ВD, then:
∠ABD = ∠ABS – ∠DBC or
∠АВD = 113º – 22º = 91º.
Answer: the value of the angle ∠АВD = 91º.