For $ABCD$ trapezoid: Let $ [FHE]=[CHB]=y $, $ [FGE]=[ADG]=x $. We have that $ [FHC] \cdot [EHB]=y^2 $ and $ [EBCF]\ge 4y $. Analoguously, $ [ADFE]\ge 4x $. So $ [ABCD]\ge 4[HFGE] $.

The proof for trapezoid is given above by mathuz. For non trapezoid it is enough to observe the limit case, when $E=A$, $F=D$, and $H$ is the intersection of diagonals of $ABCD$. In this case the statement is equivalent to $S(ADH)\leq \frac{1}{4}S(ABCD)$ which obviously couldn't be true for any quadrilateral.