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

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 ,$\\\therefore \angle AOB+\angle COD=360-1/2(\angle A+\angle B+\angle C+\angle D)=360-(1/2)360=180$

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