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Math Help - difficult hsc geometry

  1. #1
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    difficult hsc geometry

    In triangle ABC, angle BAC = 80 degrees and Q is the point on BC such that AQ bisects angle BAC.

    Point P also lies on AB such that angle QPC = 40 degrees.
    P is distinct from A.

    Prove that PQ = QC.
    Last edited by the undertaker; August 21st 2010 at 08:41 PM. Reason: ERROR
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  2. #2
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    Quote Originally Posted by the undertaker View Post
    In triangle ABC, angle BAC = 80 degrees and Q is the point on BC such that AQ bisects angle BAC.

    Point B also lies on AB such that angle QPC = 40 degrees.
    P is distinct from A.

    Prove that PQ = QC.
    angle QPC=angle QAC=40

    This shows that the quadrilateral APQC is bounded by the circle and with the use of one of the properties of circle, you will get your prove.
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  3. #3
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    CPQ and CAQ are 2 angles drawn on the same base and on the same site
    and both are equal then they lie on the same circle and the base is a chord
    on the circle, this means that APQC is a quadrilateral pie

    In quadrilateral pie each opposite angles have the sum of 180 degrees.

    angle PAC + angle PQC = 180 degrees
    80 + angles PQC = 180
    angle PQC = 180 - 80 = 100 degrees

    angle QCP = 180 - (100 + 40) = 40 deg.
    then PQ = QC


    Faraj Razem
    Last edited by razemsoft21; August 22nd 2010 at 03:45 AM. Reason: typing error
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