<s>[INST] Vector triangle lines intersection proof 
In this triangle, the distance between $M$ and $A$ is twice the distance between $M$ and $B$. And the measure of AN is the third part of the measure of $CN$. Write $X$ in function of $A$, $AB$ and $AC$.
I don't know how to write vector notation, but A is a point, and the rest are vectors. So I started trying to write $X$ like this:
$X = A + AM + MX$
And I know that $AM = 2MB$ so $AM = \frac{2}{3}AB$, then:
$X = A + \frac{2}{3}AB + MX$
I just need to get rid of the $MX$ but when I try to write it in function of other things, I Always get some that can't be written in function of $A$, $AB$ and $AC$.
The answer is $X = A + \frac{3}{5}AB + \frac{1}{10}AC$
Thanks!
 [/INST] Note that:
$$X= A + \frac{2}{3} AB + MX \quad (1)$$
$$X= A + \frac{1}{4} AC + NX \quad (2)$$
$$MX = \lambda MC = \lambda(MA +AC)= \lambda (- \frac{2}{3} AB +AC) \quad (3)$$
and
$$NX = \mu NB = \mu(NA +AB)= \mu (- \frac{1}{4} AC + AB). \quad (4)$$
Substituting $(3)$ in $(1)$ and $(4)$ in $(2)$ we get:
$$X= A + ( \frac{2}{3}- \frac{2 \lambda}{3})AB + \lambda AC \quad (5)$$
and
$$X= A + ( \frac{1}{4} - \frac{\mu}{4})AC + \mu AB \quad (6).$$
From $(5)$ and $(6)$ you get your answer.
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