Prove that if a circle can be described near a trapezoid, then this trapezoid is isosceles.

Since the AВСD quadrangle is inscribed in a circle, the sum of the opposite angles of such a quadrangle is 1800.

Then:

The angle ∠А + ∠С = 180, which means the angle ∠С = (180 – ∠А). (1)

The angle ∠В + ∠Д = 180, which means the angle ∠В = (180 – ∠Д). (2)

Since, by condition, AVSD is a trapezoid, the sum of its angles at the lateral sides is 180.

Then:

Angle ∠А + ∠В = 180, angle ∠А = (180 – ∠В). (3)

Angle ∠С + ∠Д = 180, angle ∠С = (180 – ∠Д). (4)

From 1 and 3 it follows that the angle С = А, from 2 and 4 it follows that ∠В = ∠С.

Since the angles of a trapezium at its base are equal, then such a trapezoid is isosceles, which was required to prove.