Denote the intersection of the lines $EF$ and $AG$ by $I$ and the intersection of the lines $EF$ and $AH$ by $J$.
Now, $EF$ is parallel to $BD$ and $AH$ is parallel to $CE$, so the quadrilateral $EGHJ$ is a parallelogram. Furthermore, $AG$ and $CF$ are parallel, so $FIGH$ is also a parallelogram. Hence, $|FH| = |IG|$, $|HJ| = |GE|$ and $\angle FHJ = \angle IGE$, which means that the triangles $FHW$ and $IGE$ are congruent. Thus, $|FJ| = |IE|$.

Since $EF$ is parallel to $BD$, there are three pairs of similar triangles: $EAF$ and $BAD$, $EAI$ and $BAG$, $JAF$ and $HAD$. Hence, $\frac{|EA|}{|BA|} = \frac{|FA|}{|DA|} \cdot \frac{|EA|}{|BA|} = \frac{|EI|}{|BG|}$ and $\frac{|FA|}{|DA|} = \frac{|JF|}{|DH|}$. We see that $\frac{|EI|}{|BG|} = \frac{|JF|}{|DH|}$ and together with $|FJ| = |IE|$ this implies $|BG| = |DH|$.

Let $S$ be the midpoint of the segment $AC$. Since $AGCH$ is a parallelogram, $S$ is also the midpoint of the segment $GH$. The equality $|BG| = |DH|$ implies that $S$ is the midpoint of $BD$ as well. The segments $AC$ and $BD$ bisect one another, so $ABCD$ is a parallelogram.