geometry

In triangle ABC, E is on AB such that AE:EB = 3:2, and D is on AC such that AD:DC = 1:4.  
F is the intersection of BD and CE.  
Find EF:FC.

\(\begin{array}{lclcll} \text{Let $\vec{AB}=\vec{b}$ } && \text{Let $\vec{AC}=\vec{c}$ }\\\\ \text{Let $\dfrac{AE}{AB}=\lambda=\dfrac{3}{5} $ } && \text{Let $1-\lambda=\dfrac{2}{5} $ }\\ \text{Let $\dfrac{AD}{AC}=\mu=\dfrac{1}{5} $ } && \text{Let $1-\mu=\dfrac{4}{5} $ } \\ \text{Let $\dfrac{EF}{EC}=x $ } && \text{Let $\dfrac{FC}{EC}=1-x $ }\\ \text{Let $\dfrac{EF}{FC}=\frac{x}{1-x} $ }\\\\ \text{Let $\dfrac{DF}{DB}=\eta $ }\\\\ \text{Let $\vec{AE}=\lambda\vec{b}$ } && \text{Let $\vec{AD}=\mu\vec{c}$ }\\\\ \text{Let $\vec{EC}=\vec{AC}-\vec{AE}=\vec{c}-\lambda\vec{b}$ } &&\text{Let $\vec{DB}=\vec{AB}-\vec{AD}=\vec{b}-\mu\vec{c}$ } \\\\ \text{Let $\vec{EF}=x\vec{EC}=x(\vec{c}-\lambda\vec{b})$ } &&\text{Let $\vec{DF}=\eta\vec{DB}=\eta(\vec{b}-\mu\vec{c})$ } \\ \end{array} \)

\(\begin{array}{|rcll|} \hline \vec{AE} + \vec{EF} &=& \vec{AD} + \vec{DF} \\ \lambda\vec{b} + x(\vec{c}-\lambda\vec{b}) &=& \mu\vec{c} + \eta(\vec{b}-\mu\vec{c}) \\ \lambda\vec{b} + x\vec{c}-x\lambda\vec{b} &=& \mu\vec{c} + \eta\vec{b}-\eta\mu\vec{c} \\ \vec{b}(\underbrace{\lambda-x\lambda- \eta}_{=0} ) &=& \vec{c} (\underbrace{\mu-\eta\mu- x}_{=0}) \\ \hline \lambda-x\lambda- \eta &=& 0 \\ \eta &=& \lambda-x\lambda \\ \mathbf{ \eta } &=& \mathbf{ \lambda(1-x) } \\ \hline \mu-\eta\mu- x &=& 0 \\ x &=& \mu-\eta\mu \\ x &=& \mu-\lambda(1-x)\mu \\ x &=& \mu-\lambda\mu(1-x) \\ x &=& \mu-\lambda\mu + \lambda\mu x \\ x - \lambda\mu x &=& \mu-\lambda\mu \\ x (1- \lambda\mu) &=& \mu(1-\lambda) \\ \mathbf{x} &=& \mathbf{\dfrac{ \mu(1-\lambda)}{1- \lambda\mu} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{\dfrac{ \mu(1-\lambda)}{1- \lambda\mu} } \\\\ x &=& \dfrac{ \dfrac{1}{5}* \dfrac{2}{5}}{1- \dfrac{3}{5}*\dfrac{1}{5} } \\\\ x &=& \dfrac{ \dfrac{2}{25}}{1- \dfrac{3}{25} } \\\\ x &=& \dfrac{ \dfrac{2}{25}}{ \dfrac{22}{25} } \\\\ x &=& \dfrac{2}{22} \\\\ \mathbf{x} &=& \mathbf{\dfrac{1}{11}} \\\\ \mathbf{1-x} = 1-\dfrac{1}{11} &=& \mathbf{\dfrac{10}{11}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{EF}{FC} &=& \dfrac{x}{1-x} \\\\ \dfrac{EF}{FC} &=&\dfrac{ \dfrac{1}{11} } { \dfrac{10}{11} } \\\\ \mathbf{\dfrac{EF}{FC}} &=& \mathbf{\dfrac{1}{10}} \\ \hline \end{array}\)