If A, B, C, D are four points such that BAC = 30° and BDC = 60°, then D is the center of the circle through A, B and C.

If AOB is a diameter of a circle and C is a point on the circle, then AC² + BC² = AB²

ABCD is a cyclic quadrilateral such that A = 90°, B = 70°, C = 95° and D = 105°.

If A, B, C and D are four points such that BAC = 45° and BDC = 45°, then A, B, C, D are con cyclic.

In Fig. 10.10, if AOB is a diameter and ADC = 120°, then CAB = 30°