Let $A_1$ and $D_1$ be points on the lines $AB$ and $CD$, respectively, such that $M\in\ A_1D_1$ and $A_1D_1\| BC$. Assume without loss of generality that $A$ lies on the segment $A_1B$. Let $l$ be the line parallel to $BM$ and $CD$ through $A_1$ and denote $X=l\cap AB$. Then, $S_{MD_1B}=S_{MA_1X}\geqslant S_{MA_1A}$. Moreover, because of the parallel lines, we have $S_{BCM}=S_{CD_1M}=S_{A_1BM}$. Consequently, $\left.S_{ABCD}-3S_{BCM}=S_{ABM}+S_{BCM}+S_{CD_1M}+S_{MD_1B}-S_{A_1BM}-S_{BCM}-S_{MCD_1}=S_{MD_1B}-S_{MA_1A}\geqslant 0\right.$, as desired. Equality holds if and only if $AD$ is identical to $A_1D_1$, i.e. iff $AD\|BC$.