In triangle PQR, the line ST is drawn parallel to QR so that PS= 3SQ. prove that PT = 3TR
1. Let $\displaystyle PT = k \cdot TR$
2. You are dealing with similar triangles. So use proportion:
$\displaystyle \dfrac{PS}{PS + SQ} = \dfrac{PT}{PT + TR} $
3. Substitute
$\displaystyle PS= 3 \cdot SQ$ ...... and ...... $\displaystyle PT = k \cdot TR$
$\displaystyle \dfrac{3 \cdot SQ}{3 \cdot SQ + SQ} = \dfrac{k \cdot TR}{k \cdot TR + TR} $
$\displaystyle \dfrac{3 \cdot SQ}{4 \cdot SQ } = \dfrac{k \cdot TR}{(k+1) \cdot TR } $
$\displaystyle \dfrac{3}{4 } = \dfrac{k}{(k+1) } ~\implies~3(k+1) = 4k ~\implies~\boxed{3=k}$