Let $ABCD$ be a quadrilateral, let $O$ be the intersection point of diagonals $AC$ and $BD$, and let $P$ be the intersection point of sides $AB$ and $CD$. Consider the parallelograms $AODE$ and $BOCF$. Prove that $E, F$ and $P$ are collinear.
After applying a homothety with center $O$ and factor $1/2$, the statement is equivalent to the collinearity of the midpoints of $AD,BC,PO$. This is the well-known Newton-Gauss line.
Let $FC\cap{ED}=T$ , $FB\cap{EA}=S$
By the converse of Desargues' $\triangle ASB$ and $\triangle DTC$ are perspective . And so it suffices to prove that $AD ,TS , BC$ concur
I'll complete the solution later