Question

# Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angle at the center of the circle.

Solution

## Data : Circle withe centre 'O' is circumscribed in a quadrilateral ABCD To Prove : $$\angle AOD + \angle BOC = 180^{\circ}$$Sides of the quadrilateral AB , BC , CD and DA touches at the points P, Q , R and S respectively OA . OB , OC , OD and OP , OQ , OR , OS are joined OA bisects $$\angle POS$$$$\angle 1 = \angle 2$$$$\angle 3 = \angle 4$$$$\angle 5 = \angle 6$$$$\angle 7 = \angle 8$$$$2 (\angle 1 + \angle 4 + \angle 5 + \angle 8 ) = 360^{\circ}$$$$( \angle 1 + \angle 8 ) ( \angle 4 + \angle 5 = 180^{\circ}$$$$\therefore AOD = BOD = 180^{\circ}$$$$\therefore$$ opposite sides of a  quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. Mathematics

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