In this specific example,
In $\triangle ABC$, $E$ is on $AC$ such that $AE:EC=5:1$, and $F$ is a point on $AB$ such that $AF:FB=3:2$. If $BE$ and $CF$ intersect at $M$, find the ratios $BM:ME$ and $CM:MF$.
I know this can be solved using vectors, and in this example $BM:ME=4:1$ and $CM:MF=1:2$.
However, is there a way of solving this that only uses (synthetic) geometric techniques? Will such a method work for all similar problems (i.e. the ratios of $AF:FB$ and $AE:EC$ are different).
Menelaus' theorem for triangle $\,\triangle ACF\,$ and transversal $\,E$-$M$-$B\,$ gives:
$$ \frac{AE}{EC}\cdot \frac{CM}{MF}\cdot \frac{FB}{BA} = -1 $$
Let $\,\dfrac{AE}{EC} = e\,$ and $\,\dfrac{AF}{FB}=f\,$, then the above can be written as $\,e \cdot \dfrac{CM}{MF} \cdot \dfrac{-1}{f+1}=-1\,$, from which it follows that $\,\dfrac{CM}{MF}=\dfrac{f+1}{e}\,$.
The other ratio $\,\dfrac{BM}{ME}\,$ can be derived in a similar way from $\,\triangle ABE\,$ and transversal $\,F$-$M$-$C\,$.