ALL the sides given are vectors; for example, $AB$ is a vector that goes from $A$ to $B$, and of course $AB = -BA$.
Question states: "In triangle $ABC$, points $E$ and $F$ divide sides $AC$ and $AB$ respectively so that $AE/EC = 4$ and $AF/FB = 1$. Suppose $D$ is a point on side $BC$, let $G$ be the intersection of $EF$ and $AD$, and suppose $D$ is situated so that $AG/GD = 3/2$. Find the ratio $BD/DC$."
Please draw a figure in order to better understand the problem; I tried finding a suitable one online but couldn't.
For this problem, I was stuck precisely because I had way too many equations. Here are just few of the equations that I came up with:
$$CG = \frac{3}{5}CD + \frac{2}{5}CA = \frac{3}{5}CD + 2CE$$ $$BG - \frac{3}{5}BD+\frac{2}{5}BA = \frac{3}{5}BD+\frac{4}{5}BF$$ $$CF = \frac{1}{2}CB+\frac{1}{2}CB+\frac{5}{2}CE$$ $$BE = \frac{4}{5}BC+\frac{1}{5}BA=\frac{4}{5}(BD + DC) + \frac{2}{5}BF$$ $$BG-CG = BC = BD+DC$$ $$BC=BA+AC=2BF+5EC$$
I am not sure which equations are relevant to solving problems and which ones are not. Just too many equations.
Hint: express everything in terms of $\overrightarrow{AB}, \overrightarrow{AC}$ and the unknown ratios $\lambda = \frac{GE}{FE}$, $\mu=\frac{DC}{BC}$.
The problem gives: $\;\;\overrightarrow{AE} = \frac{4}{5}\overrightarrow{AC}\;$, $\;\;\overrightarrow{AF}=\frac{1}{2}\overrightarrow{AB}\;$, $\;\;\overrightarrow{AG}=\frac{3}{5}\overrightarrow{AD}$
By construction: $$\;\;\overrightarrow{AG} = \lambda \,\overrightarrow{AF}+(1-\lambda)\,\overrightarrow{AE} = \frac{\lambda}{2} \,\overrightarrow{AB} + \frac{4(1-\lambda)}{5}\overrightarrow{AC} \\ \;\;\overrightarrow{AD}=\mu\,\overrightarrow{AB}+(1-\mu)\,\overrightarrow{AC}$$
Replacing the expressions above in $\;\overrightarrow{AG}=\frac{3}{5}\overrightarrow{AD}\;$ and identifying the coefficients of $\overrightarrow{AB}, \overrightarrow{AC}$ (since the vectors are linearly independent) gives two scalar linear equations which can be solved for $\lambda, \mu$. Then of course $\frac{BD}{DC}$ can be derived from $\mu$.