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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

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2 Answers 2

At this moment I'm only able to answer one half of your question, but I think it is worth it to post it yet. Things depend a bit on what you understand to be affine geometry, though. I for now assume that you're working with some kind of axiomatic system, including the statement by Desargues. If this is not the case, please say what axioms you're working with, so I can retry.

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.

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How about using the fact that the diagonals bisect each other? –  Mathematician Jan 29 '13 at 15:51

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.

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