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In triange ABC, D is on B so that AD:DB = 1:2, and G is on CD so that CG:GD = 3:2.  If BG intersects AC at F, find BG:GF.

 Dec 6, 2019
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In triange ABC, D is on B so that AD:DB = 1:2, and G is on CD so that CG:GD = 3:2. 

If BG intersects AC at F, find BG:GF.

 

\(\text{Let $AB=\vec{b}$} \\ \text{Let $AC=\vec{c}$} \\ \text{Let $BF=\vec{f}$} \\ \text{Let $CD=\vec{d}$} \)  

\(\begin{array}{|rcll|} \hline (1-\lambda)\vec{b}+x\vec{f}+(1-\mu)\vec{d} &=& \vec{0} \quad | \quad \vec{d} = \lambda\vec{b} - \vec{c},\ \vec{f} = u\vec{c} - \vec{b} \\ (1-\lambda)\vec{b}+x(u\vec{c} - \vec{b})+(1-\mu)(\lambda\vec{b} - \vec{c}) &=& \vec{0} \\ (1-\lambda)\vec{b}+xu\vec{c} - x\vec{b}+(1-\mu)\lambda\vec{b} - (1-\mu)\vec{c} &=& \vec{0} \\ \vec{b}\left[ (1-\lambda) -x +(1-\mu)\lambda \right] &=& \vec{c}\left[ (1-\mu)-xu \right] \\ \vec{b}\left[\underbrace{ (1-\lambda) -x +(1-\mu)\lambda} _{=0}\right] &=& \vec{c}\left[ \underbrace{ (1-\mu)-xu }_{=0}\right] \\\\ (1-\lambda) -x +(1-\mu)\lambda &=& 0 \\ 1-\lambda -x +\lambda -\mu\lambda &=& 0 \\ 1 -x -\mu\lambda &=& 0 \\ \mathbf{x} &=& \mathbf{1-\mu\lambda} \quad | \quad \mu = \dfrac{3}{5},\ \lambda=\dfrac{1}{3} \\ x &=& 1-\dfrac{3}{5}*\dfrac{1}{3} \\ x &=& 1-\dfrac{1}{5} \\ \mathbf{x} &=& \mathbf{\dfrac{4}{5}} \\\\ 1-x &=& 1- \dfrac{4}{5} \\ \mathbf{1-x} &=& \mathbf{\dfrac{1}{5}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{BG}{GF}} \\\\ &=& \dfrac{x}{1-x} \\\\ &=& \dfrac{4}{5}\above 1pt \dfrac{1}{5} \\\\ &=& \dfrac{4}{5}*\dfrac{5}{1} \\\\ &=& \mathbf{\dfrac{4}{1}} \\ \hline \end{array}\)

 

laugh

 Dec 6, 2019

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