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In the given figure, quadrilateral PQRS circumscribes the circle with centre O. Prove that the opposite sides of the quadrilateral PQRS subtend supplementary angles at the centre O.
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Solution:
Let the circle touch the sides of the quadrilateral PQRS at A, B, C and D.
Construction: Join OA, OB, OC, OD, OP, OQ, OR and OS
$\triangle ODS \cong \triangle ODS$
$\therefore \angle 1 = \angle 2$ (1 Mark)
Similarly, $\angle 4 = \angle 3$
$\angle 5 = \angle 6$
$\angle 8 = \angle 7$ (1/2 Mark)
$\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$
$2(\angle 1 + \angle 8 + \angle 4 + \angle 5) = 360^\circ$ (1/2 Mark)
$\angle SOR + \angle POQ = 180^\circ$ (1 Mark)
Similarly, $\angle POS + \angle ROQ = 180^\circ$ (1/2 Mark)
Let the circle touch the sides of the quadrilateral PQRS at A, B, C and D.
Construction: Join OA, OB, OC, OD, OP, OQ, OR and OS
$\triangle ODS \cong \triangle ODS$
$\therefore \angle 1 = \angle 2$ (1 Mark)
Similarly, $\angle 4 = \angle 3$
$\angle 5 = \angle 6$
$\angle 8 = \angle 7$ (1/2 Mark)
$\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$
$2(\angle 1 + \angle 8 + \angle 4 + \angle 5) = 360^\circ$ (1/2 Mark)
$\angle SOR + \angle POQ = 180^\circ$ (1 Mark)
Similarly, $\angle POS + \angle ROQ = 180^\circ$ (1/2 Mark)