Q. 6 B5.0( 2 Votes )

In the picture, t

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.


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


But PQRS is a cyclic quadrilateral

SQP + PRS = 180°

BAP + PRS = 180°

Now, PRS + CRS = 180° [linear pair]

BAP + 180° CRS = 180°


But CDSR is also cyclic

CRS + CDS = 180°

BAP + CDS = 180°


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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