  Question

A circle touches all the sides of a quadrilateral. Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.

Solution

Given : A quadrilateral $$PQRS$$ circumscribes a circle with center $$O$$. Sides of quadrilateral $$PQ, QR, RS$$ and $$SP$$ touches the circle at point $$L, M, N, T$$ respectively.To prove : $$\angle POQ + \angle SOR = 180^{\circ}$$and $$\angle SOP + \angle ROQ = 180^{\circ}$$Construction : Join $$P, Q, R, L, M, N$$ and $$T$$ with center $$O$$ of circle,Proof : Since, $$OL, OM, ON$$ and $$OT$$ are radius of circle and $$QL, MQ, RN$$ and $$ST$$ are tangents of circle. So$$QL \perp OL \perp QM \perp OM, RN \perp ON$$ and $$ST \perp OT$$ Now in right angled $$\Delta OMQ$$ and right $$\Delta OLQ$$$$\angle OMQ = \angle OLQ$$ ( each  $$90^{\circ}$$)hypotenuse $$OQ$$ = hypotenuse $$OQ$$ (common side)and $$OM = OL$$ (equal radii of circle)$$\therefore OMQ = OLQ$$ (By $$RHS$$ Congruence)$$\Rightarrow \angle 3 = \angle 2 (CPCT)$$Similarly $$\angle 4 = \angle 5$$$$\angle 6 = \angle 7$$ and $$\angle 8 = \angle 1$$$$\because$$ Sum of all angles made on point $$O$$ of center of circle $$= 360^{\circ}$$$$\therefore \angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^{\circ}$$$$\Rightarrow \angle 1 + \angle 2 + \angle 2 + \angle 5 + \angle 5 + \angle 6 + \angle 6 + \angle 1 = 360^{\circ}$$$$\Rightarrow 2(\angle 1 + \angle 2 + \angle 5 + \angle 6) = 360^{\circ}$$$$\Rightarrow (\angle 1 + \angle 2) + (\angle 5 + \angle 6) = 180^{\circ}$$$$\angle POQ + \angle SOR = 180^{\circ}$$$$[\because \angle 1 + \angle 2 = \angle POQ$$ and $$\angle 5 + \angle 6 = \angle SOR]$$$$\angle SOP + \angle ROQ = 180^{\circ}$$So, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of circle. Mathematics

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