In a general triangle, the point $D$ is defined such that $\vec{AD} = \frac{1}{3} \vec{AC}$, and $E$ such that $\vec{CE} = \frac{1}{4} \vec{CB}$. $F$ is the intersection of $\vec{AE}$ and $\vec{BD}$.
The task is to find $\vec{BF}$ as linear combination of $\vec{AB}$ and $\vec{AC}$.
I'm aware that finding the $F$ would be straightforward by building a system of equations based on those two straights and solving them parameter-based. However, that question specifically mentions linear combinations and is targeted at students who don't necessarly know about solving such equation systems yet.
Is there a way to solve this problem by using linear combinations and geometric rules only?
Quick sketch of the problem:
Let $I$ on $CA$ so that $EI\parallel BD$, so $CI:CD=1:4$. Therefore we have
$$\vec{EI} = \frac14 \vec{BD}.\tag{1}$$
On the other hand, $CI:CD=1:4$ implies $CI = AC/6$, and so $AD:AI = 2:5$. Thus
$$ \vec{DF} = \frac25\vec{EI}.\tag2$$
From (1) and (2), $$\vec{DF} = \frac1{10}\vec{BD},$$ and so $$\vec{BF} = \frac9{10}\vec{BD} = \frac{9}{10}(\vec{AD} - \vec{AB}).$$ Replace $\vec{AD} = \frac13\vec{AC}$ and we are done.