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Math Help - Parallelogram

  1. #1
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    Parallelogram

    A Parallelogram ABCD is given. You draw a line which is parallel to AB and crosses the sides [tex]AD[math and  BC in S and Q.
    Another line which is parallel to BC crosses the sides  AB and DC in  P and R. Show that the line that goes through SR and the line that goes through PQ intersect with the line that goes through AC in point T. When does that happen & when are those 3 lines parallel..
    (see attachment).

    -------------

    Here's how far i've managed to get..

    \vec{AB} = \vec{a}
    \vec{AD} = \vec{BC} = \vec{b}

     \vec{AP} = \alpha\vec{AB} = \alpha\vec{a}
     \vec{BQ} = \vec{AS} = \beta\vec{AD} = \beta\vec{b}
     \vec{AT} = \gamma\vec{AC} = \vec{AP} + \delta\vec{PQ}

    \vec{PQ} = \vec{PB} + \vec{BQ} = (1 - \alpha)\vec{a} + \beta\vec{b} \Rightarrow
    \Rightarrow \vec{AT} = \gamma(\vec{a} + \vec{b}) = \alpha\vec{a} + \delta((1 - \alpha)\vec{a} + \beta\vec{b})

    From that we can see:

     \vec{a} (\gamma - \alpha - \delta + \delta\alpha) + \vec{b}(\gamma - \delta\beta) = 0

    \gamma - \alpha - \delta + \delta\alpha = 0
     \gamma - \delta\beta = 0 \Rightarrow \gamma = \delta\beta

    \delta\beta - \alpha - \delta + \delta\alpha = 0
    \delta(\beta - 1 + alpha) = 0
     \delta = \frac{\alpha}{\beta - 1 + \alpha}

    Now we know when the lines are parallel.. If we get 0 in the denominator then  \delta will have an indefinite value -> meaning the lines will never intersect..

     \vec{AC} \parallel \vec{PQ} \Longleftrightarrow \alpha + \beta = 1

    ater that i'm lost.. what do i need to do now and how?
    Attached Thumbnails Attached Thumbnails Parallelogram-parallelogram.jpg  
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