Problem - 3107
As shown, $E, F, G, H$ are midpoints of the four sides. If $AB\parallel A'B', BC\parallel B'C',$ and $CD\parallel C'D'$, show that $AD\parallel A'D'$.
Connecting $AA'$, $BB'$, $CC'$ and $DD'$. We are going to show that $S_{\triangle{AA'G}}=S_{\triangle{DD'G}}$. If this holds, then they must have equal altitudes because their bases are equal. Then, $AD\parallel A'D'$.
Firstly, the areas of the four corner triangles are equal:$S_{\triangle{AGH}}=S_{\triangle{BHE}}=S_{\triangle{CEF}}=S_{\triangle{DFG}}=\frac{1}{4}S_{ABCDE}$.
Then break them each into two smaller triangles and note $(AA'H) = (BB'H)$ etc will lead to the final result.