Parallelogram

**metlx** · November 8th 2009, 01:01 PM

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).

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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?

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