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. 

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 $BC$.
