Note that $ABCD$ is a parallelogram, then $AD = BC$. Hence, we have
$\begin{array}{rcl}
BE : EC : AD & = & BE : EC : BC \\
& = & 5 : 3 : 5 + 3 \\
& = & 5 : 3 : 8
\end{array}$
Since $\Delta BEF \sim \Delta DAF$, then we have
$\begin{array}{rclcl}
\dfrac{EF}{AF} & = & \dfrac{BF}{DF} & = & \dfrac{BE}{DA} \\
\dfrac{EF}{AF} & = & \dfrac{BF}{DF} & = & \dfrac{5}{8}
\end{array}$
Consider $\Delta ABF$ and $\Delta AFD$. $\Delta ABF$ and $\Delta AFD$ have the same height with respect to the base $BF$ and $FD$ respectively. Hence, we have
$\begin{array}{rcl}
\dfrac{\text{area of $\Delta ABF$}}{\text{area of $\Delta ADF$}} & = & \dfrac{BF}{FD} \\
\dfrac{120}{\text{area of $\Delta ADF$}} & = & \dfrac{5}{8} \\
\text{area of $\Delta ADF$} & = & 192 \text{ cm}^2
\end{array}$
Consider $\Delta ABF$ and $\Delta BEF$. $\Delta ABF$ and $\Delta BEF$ have the same height with respect to the base $AF$ and $FE$ respectively. Hence, we have
$\begin{array}{rcl}
\dfrac{\text{area of $\Delta ABF$}}{\text{area of $\Delta BEF$}} & = & \dfrac{AF}{FE} \\
\dfrac{120}{\text{area of $\Delta BEF$}} & = & \dfrac{8}{5} \\
\text{area of $\Delta BEF$} & = & 75 \text{ cm}^2
\end{array}$
Since $ABCD$ is a parallelogram with diagonal $BD$, then we have
$\begin{array}{rcl}
\text{area of $\Delta ABD$} & = & \text{area of $\Delta BCD$} \\
\text{area of $\Delta ABF$} + \text{area of $\Delta ADF$} & = & \text{area of $\Delta BEF$} + \text{area of $CDFE$} \\
120 + 192 & = & 75 + \text{area of $CDFE$} \\
\text{area of $CDFE$} & = & 237 \text{ cm}^2
\end{array}$
