Since $\Delta EBF \sim \Delta DAE$, then we have
$\begin{array}{rcll}
\dfrac{EF}{DE} & = & \dfrac{EB}{DA} & \text{(corr. sides, $\sim \Delta$)} \\
\dfrac{EF}{DE} & = & \dfrac{EB}{AE+EB} \\
\dfrac{EF}{DE} & = & \dfrac{EB}{3EB+EB} \\
\dfrac{EF}{DE} & = & \dfrac{1}{4}
\end{array}$
Let $EF=k$ and $DE=4k$, where $k$ is a positive constant.
Consider $\Delta DEF$, we have
$\begin{array}{rcl}
\dfrac{1}{2} \times EF \times DE & = & 25 \\
\dfrac{1}{2} \times k \times 4k & = & 25 \\
k^2 & = & \dfrac{25}{2} \\
k & = & \dfrac{5}{\sqrt{2}}
\end{array}$
Therefore, $EF=\dfrac{5}{\sqrt{2}}\text{ cm}$ and $DE = \dfrac{20}{\sqrt{2}}\text{ cm}$.
In $\Delta ADE$, since $AD=4EB$ and $AE=3EB$, then we have $AD:AE = 4:3$. Let $AD=4m$ and $AE=3m$, where $m$ is a positive constant.
$\begin{array}{rcll}
AD^2 +AE^2 & = & DE^2 & \text{(Pyth. Thm.)} \\
(4m)^2+(3m)^2 & = & \left(\dfrac{20}{\sqrt{2}}\right)^2 \\
25m^2 & = & 200 \\
m^2 & = & 8 \\
m& = & \sqrt{8}
\end{array}$
Therefore, $AD=4\sqrt{8}\text{ cm}$ and $AE=3\sqrt{8}\text{ cm}$.
$\begin{array}{rcl}
\dfrac{FB}{AE} & = & \dfrac{1}{4} \\
\dfrac{FB}{3\sqrt{8}} & = & \dfrac{1}{4} \\
FB & = & \dfrac{3\sqrt{8}}{4}\text{ cm}
\end{array}$
Therefore, we have
$\begin{array}{rcll}
CF & = & BC – BF \\
CF & = & AD – BF & \text{(properties of square)} \\
CF & = & 4\sqrt{8} -\dfrac{3\sqrt{8}}{4} \\
CF & = & \dfrac{13\sqrt{8}}{4} \text{ cm}
\end{array}$
Therefore, the area of $\Delta CDF$
$\begin{array}{cl}
= & \dfrac{1}{2} \times CD \times CF \\
= & \dfrac{1}{2} \times AD \times CF & \text{(properties of square)} \\
= & \dfrac{1}{2} \times 4 \sqrt{8} \times \dfrac{13\sqrt{8}}{4} \\
= & 52\text{ cm}^2
\end{array}$
