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.


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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