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Question

# A circle touches the sides of a quadrilatieral ABCD at P, Q, R, S respectively. The angles subtended at the centre by a pair of opposite sides have theirs sum as:

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Solution

## ABCD is a quadrilateral circumscribing the circle with a center O. ABCD touches the circle at points, P,Q,R and S. In △AOP and △AOS.AP=AS(Length of tangents draw from external point to a circle are equal)AO=AO(Common) OP=OS(both radius)∴△AOP≅△AOS(SSS congruence rule)∠AOP=∠AOS(CPCT) Now lets number the angles for it to look easier. Q from the above conclusion, we can prove.∠2=∠3⟶(1)∠4=∠5⟶(2)∠6=∠7⟶(3) Now,∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8=360°(sum of circle round a point is 360°)∠1+∠2+∠2+∠5+∠5+∠6+∠6+∠1=360°(from (1),(2),&(3))2(∠1+∠2+∠5+∠6)=360°(∠1+∠2)+(∠1+∠2)=180∠AOB+∠COD=180 Thus, both the angles are supplementary. similarly, we can prove∠BOC+∠AOD=180. Hence proved

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