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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  • 4 years ago
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    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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