Given a Pyramid $ABCDS$ with Parallelogram $ABCD$ as a base.
The diagonals of a parallelogram intersect in point $O$.
$\vec{SA}=u,\: \vec{SB}=v,\:\vec{SC}=w,\:\vec{SF}=k\cdot\vec{SD} ,\:\vec{SE}=t\cdot\vec{SO},\: SE=EO$.
Prove that point $F$ is dividing $SD$ in ratio of $SF:FD=1:2$.
$B,E,F$ is on the same line.
I tried analytic approach but got stuck
Thanks.