vector , geometry

**Tweety** · December 31st 2008, 11:30 AM

In triangle PQR, the line ST is drawn parallel to QR so that PS= 3SQ. prove that PT = 3TR

**earboth** · January 1st 2009, 09:57 AM

Originally Posted by Tweety

In triangle PQR, the line ST is drawn parallel to QR so that PS= 3SQ. prove that PT = 3TR

1. Let $PT = k \cdot TR$

2. You are dealing with similar triangles. So use proportion:

$\dfrac{PS}{PS + SQ} = \dfrac{PT}{PT + TR}$

3. Substitute

$PS= 3 \cdot SQ$ ...... and ...... $PT = k \cdot TR$

$\dfrac{3 \cdot SQ}{3 \cdot SQ + SQ} = \dfrac{k \cdot TR}{k \cdot TR + TR}$

$\dfrac{3 \cdot SQ}{4 \cdot SQ } = \dfrac{k \cdot TR}{(k+1) \cdot TR }$

$\dfrac{3}{4 } = \dfrac{k}{(k+1) } ~\implies~3(k+1) = 4k ~\implies~\boxed{3=k}$

**Tweety** · January 1st 2009, 11:11 AM

thanks so much for that, I would have never figured out how to prove it!

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