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

Answers (1)

SOLUTION: We know that if quadrilateral cricumscribes a circle of respectively 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 ,\	herefore angle AOB+angle COD=360-1/2(angle A+angle B+angle C+angle D)=360-(1/2)360=180

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