Question

all sides of a quadrilateral touches the circle with O . prove that  angle AOB + angle COD = 180

Answer 1

we know that if a quadrilateral circumscribes a circle then the angle bisectors of the respective vertices of the quadrilateral are concurrent and the point of concurrency is the centre of the circle. hence angle OAB = 1/2 angle A and angle OBA = 1/2 angle B. therefore angle AOB = 180 - 1/2 (angle A+angle B). similarly you get angle COD = 180 - 1/2 (angle C + angle D). therefore angle AOB+angle COD = 360 - 1/2(angle A+angle B+angle C+angle D) = 360 - 1/2 * 360 = 360 - 180 = 180. hence proved.