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  • Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

13.   Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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Given- ABCD is a quadrilateral circumscribing a circle. P, Q, R, S are the point of contacts on sides AB, BC, CD and DA respectively.

To prove- 
 \\\angle AOB + \angle COD =180^0\\ \angle AOD + \angle BOC =180^0
Proof - 
Join OP, OQ, OR and OS
In triangle \DeltaDOS and \DeltaDOR,
OD =OD [common]
OS = OR [radii of same circle]
DR = DS [length of tangents drawn from an external point are equal ]
By SSS congruency, \DeltaDOS \cong \DeltaDOR,
and by CPCT, \angleDOS = \angleDOR
                          \angle c = \angle d.............(i)

Similarily,     
                    \\\angle a = \angle b\\ \angle e = \angle f\\ \angle g =\angle h...............(2, 3, 4)

 \therefore 2(\angle a +\angle e +\angle h+\angle d) = 360^0
\\(\angle a +\angle e) +(\angle h+\angle d) = 180^0\\ \angle AOB + \angle DOC = 180^0
SImilarily, \angle AOD + \angle BOC = 180^0

Hence proved.

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manish

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