Answer :

** Given:** In the given figure, the circles on the left and right intersect the middle circle at P, Q, R, S; the lines joining them meet the left and right circles at A, B, C, D

** To Prove:** ABCD is cyclic

Construction: Join PQ and RS.

Proof:

As, APBQ is a cyclic quadrilateral,

∠BAP + ∠BQP = 180° [Opposite angles in a cyclic quadrilateral are supplementary]

∠BQP = 180° – ∠BAP

Now, ∠BQP + ∠SQP = 180° [linear pair]

⇒ 180° – ∠SQP = 180° – ∠BAP

⇒ ∠SQP = ∠BAP

But PQRS is a cyclic quadrilateral

⇒ ∠SQP + ∠PRS = 180°

⇒ ∠BAP + ∠PRS = 180°

Now, ∠PRS + ∠CRS = 180° [linear pair]

⇒ ∠BAP + 180° – ∠CRS = 180°

⇒ ∠BAP = ∠CRS

But CDSR is also cyclic

⇒ ∠CRS + ∠CDS = 180°

⇒ ∠BAP + ∠CDS = 180°

Or

⇒ ∠A + ∠D = 180°

⇒ ABCD is a cyclic quadrilateral

[As in a cyclic quadrilateral, sum of any pair of opposite angles is 180°].

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