Proving that $GH$ is parallel to $AD$ in the given figure 
In the above figure, $ABCD$ and $AECF$ are two parallelograms such that $EF$ is parallel to $AB$. $DHF$ and $BGE$ are straight lines intersecting $EC$ and $AF$ at $H$ and $G$ respectively. We need to prove that $GH \parallel AD$.
One approach may be by using Basic Proportionality Theorem (BPT). However, there is no triangle where it can be applied. I tried some constructions but that only made the problem more difficult. 
 A: Draw the diagonals $AC$ and $BD$ of parallelogram $ABCD$. They intersect at a point $M$ which is the midpoint of both $AC$ and $BD$. Now $AC$ is a diagonal in the parallelogram $AECF$ whose other diagonal is $EF$. Again by the same argument, the two diagonals $AC$ and $EF$ intersect at a point which is the midpoint of both $AC$ and $EF$. But point $M$ is already the midpoint of $AB$. Hence, $EF$ passes through $M$ and moreover $M$ is the midpoint of $EF$. Thus $AC, BD$ and $EF$ all intersect in a common point $M$ which is the midpoint of all three of them simultaneously. With this information at hand, look at quadrilateral $BEDF$. Its two diagonals are $EF$ and $BD$ which we already know intersect in $M$, which is a midpoint for both of them. Therefore $BEDF$ has to be parallelogram and thus $BE$ and $DF$ are parallel. This means that $GE$ and $HF$ are parallel. We already know that $EH$ and $FG$ are parallel, hence quadrilateral $GEHF$ is a parallelogram which means that $EH = FG$. We also know that $EC = FA$ because $AECF$ is a parallelogram. Therefore $HC = AG$. By assumption, $EF$ is parallel to $AB$. Therefore $\frac{EG}{GB} = \frac{FG}{GA}$. But we already know that $FG = EH$ and $GA = HC$. Therefore
$$\frac{EG}{GB} = \frac{FG}{GA} = \frac{EH}{HC}.$$ These last relations hold if and only if $GH$ is parallel to $BC$. Since $ABCD$ is a parallelogram and $BC$ is parallel to $AD$, the segment $GH$ is parallel to $AD$. 
A: Notice first of all that all lines are symmetric around the common center $O$ of parallelograms $ABCD$ and $AECF$. It follows that $EHFG$ is a parallelogram and $EO=FO$.
Produce $GH$ to meet $AB$ at $M$. By similar triangles one has:
$$
EO:BM=OG:GM=FO:AM.
$$
Hence $AM=BM$ and $GH$ belongs to line $OM$, connecting the midpoint of $AB$ to the midpoint of $AC$ and therefore parallel to $BC$.
