Show that diagonals intersect at common point

Question

Given is octagon where opposite sides are equal length and parallel. Show that diagonals: $AE,DH, BF, CG$ intersects at point $S$

So I have tried to create a parallelograms $AHED$ and $BCFG$ and use that $AE=GC$

Answer 1

Show that $AHED$, $BGFC$ and $AGEC$ are parallelograms (which is quite obvious). Then their diagonals bisect each other. And the lines meet at their midpoints.

Answer 2

Since opposite sides are parallel, angles $BAH$ and $DEF$ are equal. Similarly angle $AHG$ and $EDC$ are equal. Also, lengths of parallel sides are equal. Now consider the line $GC$. It forms two pentagons $GHABC$ and $CDEFG$ which are equal for the reasons mentioned above.Since $S$ is the point at which line $AE$ crosses $GC$, it would be the midpoint of $AE$. Similarly using pentagons $AHBFE$ and $EDCBA$, it can be proved that the point at which $GC$ crosses $AE$ will be the midpoint of $GC$. Taking similar pairs such as BF and GC etc it can be proved that midpoint of all diagonals lie at the same point. And it is $S$