A Geometry problem 2 Given a parallelogram ABCD. A line, parallel to BC crosses AB and CD in points E and F respectively, a line parallel to AB crosses Bc and DA in points G and H respectively. Prove that EH, GF, and BD are either parallel or intersect in a single point.  I've tried to find relationships between EH and GF.How to solve this one?
 A: $\def\vec#1{\overrightarrow{#1}}$Consider $(A;\vec{AB},\vec{AD})$ as an Affine coordinate system. We have $B(1,0)$, $D(0,1)$, $E(e,0)$, $F(e,1)$, $H(0,h)$ and $G(1,h)$.
The lines $EH$, $EG$ and $BD$ have there equations as follows
$$\eqalignno{
BD:&\qquad x+y=1 &(1)\cr
EH:&\qquad h x+e y=eh &(2)\cr
FG:&\qquad (h-1) x+(e-1) y=eh -1&(3)
}$$
Noting that $(3)$ is equivalent $h x+e y-eh=x+y-1$, we conclude that if $BD$ and $EH$ intersect at a point then this point belongs to $FG$. On the other hand, if   $BD$ and $EH$ are parallel then $h=e$ and in this case they are parallel to $FG$, and we are done. 
A: Let $I, J$ be intersections of $(EH, BD)$ and  $(BD, GF)$ respectively. If you apply Menelaus' theorem twice, you have $\frac{DH}{HA}\frac{AE}{EB}\frac{BI}{ID}=1$ and $\frac{DF}{FC}\frac{CG}{GB}\frac{BJ}{JD} = 1$. Because $ABCD$ is a parallelogram and $EF, GH$ are parallel to each sides, $\frac{DH}{HA}=\frac{CG}{GB}$, $\frac{AE}{EB}=\frac{DF}{FC}$. Therefore, we have $\frac{BI}{ID}=\frac{BJ}{JD}$, which means that $I$ and $J$ are actually the same point. 
