Proving a Ratio With Vectors

Question

I was playing around with vectors in Geogebra and constructed a triangle which has a cool property; this is the dude in question:

In which $AE=EC$, and $\frac{BD}{DC}=\frac{2}{3}$. ($F$ is the intersection of $AD$ and $BE$.)
The property I mentioned is that I can play with the triangle's vertices (and, therefore, the sizes of its sides) as much as I want, and the ratios $\frac{AF}{FD} = \frac{5}{2}$ and $\frac{BF}{FE} = \frac{4}{3}$ remain constant.
I tried to prove to myself algebraically, although with simple vector arithmetics only, that this was indeed the case; but I'm missing something (presumably silly) and I keep getting stuck. What I tried to prove is that given the vectors $\vec{AB}$ and $\vec{AC}$ and the above information ($E$ is the midpoint of $AC$ and $\vec{BD}=\frac{2}{5}\vec{BC}$), it must be that $\frac{BF}{FE} = \frac{4}{3}$.

I tried several methods, but I keep finding myself with only one equation to describe two variables: one representing the ratio between AF and AD, and one the ratio between BF and FE.
Namely, if I let $\frac{AF}{AD}=\alpha$ and $\frac{BF}{BE}=\beta$, then:
$\vec{AB} + \beta \cdot \vec{BE} = \alpha \cdot \vec{AD}$
And then, expressing the rest of the vectors using $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} + \beta \cdot (\frac{1}{2}\vec{AC}-\vec{AB}) = \alpha \cdot (\frac{2}{5}\cdot\vec{AC}+\frac{3}{5}\cdot\vec{AB})$
And that's the only equation I could muster, having both $\alpha$ and $\beta$ as unknowns. (I did play with it by constructing a few others, for example ones using $\vec{FE}$ instead of $\vec{BF}$ or $\vec{FD}$ instead of $\vec{AF}$ or just generally expressing some vectors using others, but everything I tried got me either to the very same equation post-simplification, or to a good old true statement, usually followed by me realizing the triviality of the initial equation I built and going "DUH" at myself).

I feel like I'm missing something very obvious. Maybe it's some simple geometric reasoning? But I get the feeling algebra alone can definitely get me there, and I'm not sure which piece of information I'm neglecting to make proper use of.

TL;DR: How would you prove, specifically using vector algebra, that the ratios of $\frac{AF}{FD}$ and $\frac{BF}{FE}$ are constant given that $E$ is the midpoint of $AC$ and that $D$ is $\frac{2}{5}$ of $BC$, using vectors $\vec{AB}$ and $\vec{AC}$?

Answer 1

Continuing from your equation

\begin{align} \vec{AB} + \beta \cdot (\frac{1}{2}\vec{AC}-\vec{AB}) &= \alpha \cdot (\frac{2}{5}\cdot\vec{AC}+\frac{3}{5}\cdot\vec{AB})\\ (1-\beta)\vec{AB}+\frac12\beta\vec{AC}&=\frac25\alpha\vec{AC}+\frac35\alpha\vec{AB} \end{align}

But $\vec {AB}$ and $\vec {AC}$ points to different directions. This forces the system of equations

\begin{cases} 1-\beta=\dfrac35\alpha\\\dfrac12\beta=\dfrac25\alpha \end{cases}

Solving this gives $\alpha = \dfrac57, \beta = \dfrac47$.

This gives the ratio you want. The general case of this phenomenon is Menelaus's Theorem.