Consider $\Delta CDF$ and $\Delta CBF$. With bases $FD$ and $FB$, they have the same height. Hence, we have
$\begin{array}{rcl}
\dfrac{\text{area of }\Delta CDF}{\text{area of }\Delta CBF} & = & \dfrac{DF}{BF} \\
\text{area of }CDF & = & \dfrac{DF}{BF} \times \text{area of }\Delta CBF \ \ldots \unicode{x2460}
\end{array}$
Consider $\Delta CDF$ and $\Delta EBF$. Since they are similar, we have
$\begin{array}{rcl}
\dfrac{\text{area of }\Delta CDF}{\text{area of }\Delta EBF} & = & \dfrac{DF^2}{BF^2} \\
\text{area of } \Delta CDF & = & \dfrac{DF^2}{BF^2} \times \text{area of }\Delta EBF \ \ldots \unicode{x2461}
\end{array}$
Sub. $\unicode{x2460}$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
\dfrac{DF}{BF} \times \text{area of }\Delta CBF & = & \dfrac{DF^2}{BF^2} \times \text{area of }\Delta EBF \\
\text{area of }\Delta CBF & = & \dfrac{DF}{BF} \times \text{area of }\Delta EBF \ \ldots \unicode{x2462}
\end{array}$
Consider $\Delta DEF$ and $\Delta EBF$. With bases $DF$ and $BF$, they have the same height. Hence, we have
$\begin{array}{rcl}
\dfrac{\text{area of }\Delta DEF}{\text{area of }\Delta EBF} & = & \dfrac{DF}{BF} \ \ldots \unicode{x2463}
\end{array}$
Sub. $\unicode{x2463}$ into $\unicode{x2462}$, we have
$\begin{array}{rcl}
\text{area of }\Delta CBF & = & \dfrac{\text{area of }\Delta DEF}{\text{area of }\Delta EBF} \times \text{area of }\Delta EBF \\
\text{area of }\Delta CBF & = & \text{area of }\Delta DEF
\end{array}$
Therefore, the ratio of the $\text{area of }\Delta DEF$ to the $\text{area of }\Delta CBF$ is $1: 1$.
