Since $ABCD$ is a parallelogram, then $AB=CD$.
It is given that $AE=BE$ and $DF=FG=GC$. Therefore we have
$\begin{array}{rcl}
2BE & = & 3 GC \\
\dfrac{BE}{GC} & = & \dfrac{3}{2} \\
BE : GC & = & 3 : 2
\end{array}$
Since $\Delta HEB \sim \Delta HCG$, then we have $BE:GC=HE:HC=HB:HG=3:2$.
Consider $\Delta BCH$ and $\Delta BEH$, they have the same height. Hence, we have
$\begin{array}{rcl}
\dfrac{\text{Area of } \Delta BCH}{\text{Area of }\Delta BEH} & = & \dfrac{HC}{HE} \\
\dfrac{6}{\text{Area of }\Delta BEH} & = & \dfrac{2}{3} \\
\text{Area of }\Delta BEH & = & 9 \text{ cm}^2
\end{array}$
Consider $\Delta BCH$ and $\Delta GCH$, they have the same height. Hence, we have
$\begin{array}{rcl}
\dfrac{\text{Area of }\Delta BCH}{\text{Area of }\Delta GCH} & = & \dfrac{HB}{HG} \\
\dfrac{6}{\text{Area of }\Delta GCH} & = & \dfrac{3}{2} \\
\text{Area of }\Delta GCH & = & 4\text{ cm}^2
\end{array}$
Consider $\Delta BCE$ and $\Delta FEC$, they have the same height. Hence, we have
$\begin{array}{rcl}
\dfrac{\text{Area of }\Delta BCE}{\text{Area of }\Delta FEC} & = & \dfrac{BE}{FC} \\
\dfrac{6+9}{\text{Area of }\Delta FEC} & = & \dfrac{3}{4} \\
\text{Area of }\Delta FEC & = & 20\text{ cm}^2
\end{array}$
Therefore, the area of the quadrilateral $EFGH$
$\begin{array}{cl}
= & \text{Area of }\Delta FEC – \text{Area of }\Delta GCH \\
= & 20 – 4 \\
= & 16 \text{ cm}^2
\end{array}$
