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Question

# A circle touches the sides of a quadrilateral ABCD at the points P,Q,R and S respectively. Show that the angles subtended at the center by by a pair of opposite sides are supplementary.

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Solution

## Give : Let ABCD be the quadrilateral circumscribing the circle with centre O. ABCD touches the circle at point P,Q,R & S.To Prove : opposite angle subtend supplementary angles at centeri.e. ∠AOB+∠COD=180o& ∠AD+∠BOC=100oProof : In ΔAOP and ΔAOSAP=ASAO=AO (common)OP=OS (both radius)ΔAOP=ΔAOS (SSS congruence rule)∠AOP=∠AOS (CPCT)i.e. ∠1=∠8 ...(i)Similiarly, we have∠2=∠3 ...(ii)∠5=∠4 ...(iii)∠6=∠7 ...(iv)Now ∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8=360∠1+∠2+∠2+∠5+∠5+∠6+∠6+∠1=3602[∠1+∠2+∠5+∠6]=360o∠1+∠2+∠5+∠6=180o(∠1+∠2)+(∠5+∠6)=180o∠AOB+∠COD=180oHence both angles are supplementary.Similarly, ∠BOC+∠AOD=180oHence proved.

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