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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. 
1956260_1867368_ans_ef753803e8e24acfbf04e9fdc9af3278.png

Mathematics

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