I'm stuck in an exercise which consists in proving the following:
(We are working in an affine space $A$ which field $K$ contains $1/3$.)
Theorem: Let $l_1$ and $l_2$ be two lines that intersect at $O$. Let $p_1,q_1\in l_1$ and $p_2,q_2\in l_2$, and let $b_p$ be the centroid of $O$, $p_1$ and $p_2$, and $b_q$ be the centroid of $O$, $q_1$ and $q_2$. Then the vectors $p_1p_2$ and $q_1q_2$ are parallel if and only if the points $O$, $b_p$ and $b_q$ all lie in the same line.
I've attempted a lot of things, often working with equalities such as
$$Ob_p = \frac{1}{3} (Op_1+Op_2)$$
or $$p_1b_p =\frac{1}{3} (p_1O+p_1p_2)$$
as well as the fact that the points $O$, $b_p$ and $b_q$ are linear if and only if $Ob_p=kOb_q$ for some scalar $k\in K$, but I haven't been able to achieve much.
Let the two lines have direction vectors $v_1$ and $v_2$ in $\mathbb{R}^3$, then
$p_1 = t_1 v_1$
$q_1 = s_1 v_1$
$p_2 = t_2 v_2$
$q_2 = s_2 v_2$
$p_1 p_2 $ is parallel to $q_1 q_2$ if $p_1 p_2 \times q_1 q_2 = 0$ , that is,
$(t_2 v_2 - t_1 v_1) \times (s_2 v_2 - s_1 v_1) = 0$
Multiplying out and keeping in mind that $v_1 \times v_1 = v_2 \times v_2 = 0 $, we get,
$ (-t_2 s_1) (v_2 \times v_1) - t_1 s_2 (v_1 \times v_2) = 0 $
And from the properties of the cross product, we know that,
$v_2 \times v_1 = - v_1 \times v_2 $
Hence, the condition now becomes
$(t_2 s_1 - t_1 s_2 ) (v_1 \times v_2) = 0$
Since $v_1$ is not a multiple of $v_2$ because $l_1 $ and $l_2$ are different lines, then $(v_1 \times v_2)$ is not the zero vector, hence,
$ t_2 s_1 - t_1 s_2 = 0$
Which can rephrased as,
$ \dfrac{t_2}{t_1} = \dfrac{s_2}{s_1} $