$\Delta ABC$ is a triangle, $M$ is a point in the segment $\overrightarrow{AB}$ and $N$ is a point in the segment $\overrightarrow{AC}$, such that $\overrightarrow{MN}$ is parallel to $\overrightarrow{BC}$. Show that
$$\frac{|| \overrightarrow{AM}||}{|| \overrightarrow{AB}||}=\frac{|| \overrightarrow{AN}||}{|| \overrightarrow{AC}||}=\frac{|| \overrightarrow{MN}||}{|| \overrightarrow{BC}||}$$
Hint: Consider $r,s,t\in \mathbb{R}$ such that $\overrightarrow{AN}=r\cdot \overrightarrow{AC}$, $\overrightarrow{MN}=s\cdot \overrightarrow{BC}$ and $\overrightarrow{AM}=t\cdot \overrightarrow{AB}$ and use the fact that $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are linearly independent to prove $r=s=t$.
I've had this question on my exam and I thought that I needed to assume first that:
$$\frac{|| \overrightarrow{AM}||}{|| \overrightarrow{AB}||}=\frac{|| \overrightarrow{AN}||}{|| \overrightarrow{AC}||}=\frac{|| \overrightarrow{MN}||}{|| \overrightarrow{BC}||}$$
Because otherwise, I'd be assuming arbitrary triangles that wouldn't work in this case (I guess). But the professor said that this is what I should prove, with his comment I assumed that this should be the conclusion in the proof - the last step perhaps. But from here, I have no idea on how to proceed, I've meditated about the premise on the linear independence, but I'm out of ideas. I don't know the order in which I should assume what is given in the question in order to prove it.
Edit:(correct mistake)
$\overrightarrow{AM}+\overrightarrow{MN}+\overrightarrow{NA}=0 \\ \implies t\overrightarrow{AB}+s\overrightarrow{BC}+r\overrightarrow{CA}=0 \\ \implies t\overrightarrow{AB}+s\overrightarrow{BC}+r(-(\overrightarrow{AB}+\overrightarrow{BC}))=0 \\ \implies (t-r)\overrightarrow{AB}+(s-r)\overrightarrow{BC}=0$
$\overrightarrow{AB}$ , $\overrightarrow{BC}$ is linear independent $\implies t-r=0,s-r=0$