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Math Help - cyclic quadrilateral

  1. #1
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    cyclic quadrilateral

    PQRS is a cyclic quadrilateral where angle PQR and angle QRS are obtuse and have the same value . IF PQ=9 , QR=6 , and RS=4 , prove that if T is the point of intersection of the diagonals , then the circle that passes through R,S and T touches QR at R .
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  2. #2
    MHF Contributor red_dog's Avatar
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    I think it's something wrong with this problem.
    If \widehat{PQR}=\widehat{QRS}=\alpha then \widehat{QPS}=\widehat{RSP}=180-\alpha

    Then \widehat{QRS}+\widehat{RSP}=180\Rightarrow QR\parallel PS and the quadrilater is an isosceles trapezoid. Then PQ=RS.

    But PQ=9, \ RS=4.
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  3. #3
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    If this inscribed quadrilateral has 2 equal obtuse angles, it is most certainly an isosceles trapezoid / trapezium, and so PQ and RS is supposed to have the same length.
    Last edited by ukorov; November 21st 2009 at 10:16 AM.
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