Since $ABCD$ is a parallelogram, then $AD = BC$. Hence, we have
$\begin{array}{rcl}
BM:MC & = & 1:2 \\
BM : BC & = & 1 : 1+2 \\
& = & 1:3 \\
BM : AD & = & 1 : 3
\end{array}$
Since $\Delta BMG \sim \Delta DAG$, we have
$\begin{array}{rcl}
\dfrac{\text{area of }\Delta BMG}{\text{area of }\Delta DAG} & = & \left( \dfrac{BM}{DA} \right) ^2 \\
\dfrac{1}{\text{area of }\Delta DAG} & = & \left(\dfrac{1}{3} \right)^2 \\
\dfrac{1}{\text{area of }\Delta DAG} & = & \dfrac{1}{9} \\
\text{area of } \Delta DAG & = & 9 \text{ cm}^2
\end{array}$
Since $\Delta BMG \sim \Delta DAG$, we have
$\begin{array}{rcl}
BG:DG & = & BM:AD \\
& = & 1:3 \\
\end{array}$
Consider $\Delta ABG$ and $\Delta ADG$. Note that they have the same height with base $BG$ and $DG$ respectively. Therefore, we have
$\begin{array}{rcl}
\dfrac{\text{area of }\Delta ABG}{\text{area of }\Delta ADG} & = & \dfrac{BG}{DG} \\
\dfrac{\text{area of }\Delta ABG}{9} & = & \dfrac{1}{3} \\
\text{area of }\Delta ABG & = & 3 \text{ cm}^2
\end{array}$
Therefore, the area of $\Delta ABD$
$\begin{array}{cl}
= & \text{area of }\Delta ABG + \text{ area of } \Delta ADG \\
= & 9 + 3 \\
= & 12\text{ cm}^2
\end{array}$
Hence, the area of the parallelogram $ABCD$
$\begin{array}{cl}
= & 2 \times \text{area of } \Delta ABD \\
= & 2 \times 12 \\
= & 24 \text{ cm}^2
\end{array}$
