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Thread: Concurrency at interior point

  1. #1
    Senior Member pankaj's Avatar
    Jul 2008
    New Delhi(India)

    Concurrency at interior point

    Let $\displaystyle P$ be an interior point of $\displaystyle \triangle ABC $such that the lines $\displaystyle AA_{1},BB_{1},CC_{1}$ are concurrent at $\displaystyle P$ and the points $\displaystyle A_{1},B_{1},C_{1}$ lie on $\displaystyle BC,CA,AB$ respectively.Find using vectors or otherwise value of
    $\displaystyle \frac{PA_{1}}{AA_{1}}+\frac{PB_{1}}{BB_{1}}+\frac{ PC_{1}}{CC_{1}}$
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  2. #2
    MHF Contributor red_dog's Avatar
    Jun 2007
    Medgidia, Romania
    Let $\displaystyle PQ\perp BC, \ AD\perp BC$

    Then, $\displaystyle \frac{PA_1}{AA_1}=\frac{PQ}{AD}$

    $\displaystyle A_{\Delta PBC}=\frac{PQ\cdot BC}{2}, \ A_{\Delta ABC}=\frac{AD\cdot BC}{2}$

    $\displaystyle \frac{A_{\Delta PBC}}{A_{\Delta ABC}}=\frac{\frac{PQ\cdot BC}{2}}{\frac{AD\cdot BC}{2}}=\frac{PQ}{AD}$

    So, $\displaystyle \frac{PA_1}{AA_1}=\frac{A_{\Delta PBC}}{A_{\Delta ABC}}$

    In a similar way we have

    $\displaystyle \frac{PB_1}{BB_1}=\frac{A_{\Delta PAC}}{A_{\Delta ABC}}$ and $\displaystyle \frac{PC_1}{CC_1}=\frac{A_{\Delta PAB}}{A_{\Delta ABC}}$

    Then $\displaystyle \frac{PA_1}{AA_1}+\frac{PB_1}{BB_1}+\frac{PC_1}{CC _1}=\frac{A_{\Delta PBC}+A_{\Delta PAC}+A_{\Delta PAB}}{A_{\Delta ABC}}=\frac{A_{\Delta ABC}}{A_{\Delta ABC}}=1$
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