Let us suppose we have following parallelogram
Task is following : find $k$ for which in vector addition form
$AB+BO=k*AC$
interesting task, first of all, we have following equation from vector point of view
$AB+BC=AC$ $AD+DC=AC$
$BD=AD-AB$
from where, $AB=BD+AD$ , so if we insert, we will get
$BD+AD+BC=AC$
but how to continue? generally intersection point is centroid and is should be $1/2$, but how to get it from the equation? thanks in advance
Consider this:
From triangle law on $\triangle AOB$ and $\triangle AOD$, we have
$$k AC + lDB = AB \\ k AC - (1-l) DB = AD$$
So, adding the two,
$$(2k-1)AC + (2l-1)DB = 0$$
and since $AC, DB \neq 0$,
$$k = l = \dfrac{1}{2}$$