If ABCD is a parallelogram and M,N,P,Q are points on it sides then MNPQ is a paralellogram iff the diagonals intersect at a common point (i.e the diagonals of MNPQ and ABCD intersect at the same point). I want to prove this using affine geoometry. Please help

First of all, in order for $MNPQ$ to be a quadrilateral, each of $M$, $N$, $P$ and $Q$ has to be on a different side of $ABCD$. I'll say that $M$ is on $AB$, $N$ on $BC$, $P$ on $CD$ and $Q$ on $DA$. The statement by Desargues says that if we have seven points such that $AA'$, $BB'$ and $CC'$ intersect in $O$, $AB\parallel A'B'$, $BC\parallel B'C'$, then $AC\parallel A'C'$. We can apply this to $A=M$, $B=B$, $C=N$, $A'=P$, $B'=D$, $C'=Q$ and $O$ the intersection of all diagonals to obtain that $MN\parallel QP$. The same can be done for the other two sides of $MNPQ$ to obtain that it's a parallellogram.
For the converse direction, I think you need to assume that the diagonal $NQ$ does not contain the intersection point $O$ of the diagonals of $ABCD$ and arrive at some contradiction of the sort that there are two distinct lines through one point, parallel to the same line - which is impossible. I'm not yet entirely sure how to arrive here, though.
You may assume $$A=(0,0),\quad B=(1,0),\quad C=(1,1),\quad D=(0,1),$$ and $$M=(\mu,0),\quad N=(1,\nu),\quad P=(1-\rho,1),\quad Q=(0,1-\sigma)\ .$$ The quadrangle $MNPQ$ is a parallelogram iff $\vec{MN}=\vec{QP}$, i.e., iff $N-M=P-Q$. This is equivalent with $$\mu=\rho\quad\wedge\quad \nu=\sigma\ .$$ If these conditions are fulfilled then the diagonals of $MNPQ$ intersect at $\bigl({1\over2},{1\over2}\bigr)$, and conversely.