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Math Help - another circle geometry question~~~ really urgent

  1. #1
    shosho
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    another circle geometry question~~~ really urgent

    in a traingle ABC, <BAC=80 and Q is the point on BC such that AQ bisects the angle BAC. There is also a point P on AB such that <QPC=40 and P is distinct from A. Prove that PQ=QC

    any help would be awesome!!
    thanx in advnace
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  2. #2
    MHF Contributor red_dog's Avatar
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    We have \widehat{QPC}=\widehat{QAC}, then the quadrilater ACQP can be inscribed in a circle.
    Yields \widehat{PAQ}=\widehat{PCQ}=\widehat{QPC}, so the triangle PQC is isosceles, then PQ=QC
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