The center of the circle described around the trapezoid belongs to the larger base.

The center of the circle described around the trapezoid belongs to the larger base. Find the corners of the trapezoid if the bases are 1: 2.

Since the larger base passes through the center of the circle, the BP is its diameter. Since by condition, AD = 2 * BC, then BC, in length is equal to the radius of the circle.
Let us draw from the center of the circle O the radii OB and OС. Then OB = OC = BC = R, and therefore the triangle OBC is equilateral, in which all internal angles are equal to 600.
Triangle AOB is isosceles, since OA = OB = R. Angle AOB = OBC as criss-crossing angles at the intersection of parallel straight lines AD and BC secant OB. Then the angle AOB = 600, and since the triangle AOB is isosceles, the angle OAB = OBA = (180 – 60) / 2 = 60. Similarly, the angle ODС = 60.
Since the sum of the angles at the lateral sides of the trapezoid is 180, the angle ABC = ВСD = 180 – 60 = 120.
Answer: The angles of the trapezoid are 60, 120, 60, 120.