# 2 circles intersect at 2 pts A&B. PQ is chord of 1st circle,RS chord of 2nd s.t. they intersect at pt E on seg AB. Show quad PRQS is cyclic?

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- LearnerLv 74 years agoFavorite Answer
i) AB & PQ are both chords of first circle intersect at E.

So by two chords intersecting theorem of a circle, PE*EQ = AE*EB ------ (1)

ii) Similarly, from circle (2), we have, RE*ES = AE*EB -------- (2)

iii) Thus from (1) & (2): PE*EQ = RE*ES

==> PE/ES = RE/EQ

Also, <PES = <REQ [Vertically opposite angles]

So ΔPES is similar to ΔREQ [SAS similarity axiom]

==> <SPE = <QRE

As these two are angles of the same segment SQ and being equal,

Points P,R,Q,S lie on the same circle,

So PRQS is a cyclic quadrilateral.

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