In trapezoid ABCD, diagonal BD is perpendicular to side AB, BC = CD, angle A = 50 degrees, find the corners of the trapezoid.

Consider a triangle AED, in which, by condition, the angle ABC = 90, and the angle ВAD = 50, then the angle ADВ = 180 – 90 – 50 = 40.
In the ВСD triangle, by condition, the sides of the ВС and SD are equal, therefore, the triangle is equilateral, and then the angle of ВСD = ВDС.
Angle СВD = ВDA as criss-crossing angles at the intersection of parallel straight lines AD and ВС of secant ВD, then the angle СВD = СDВ = AED = 40. Then the angle ABС= 40 + 40 = 80, and the angle ABC = 90 + 40 = 130.
ВСD angle = 180 – 40 – 40 = 100.
Answer: The angles of the trapezoid are equal to ВAD = 50, ABC = 130, ВСD = 100, СDA = 80.