In the figure below, $D$ is the midpoint of side $A B, C E: D E=5: 3 \text{ and } B F: E F=1: 3$. If the area of triangle $A B C$ is $192 \text{ cm}^2$, determine the area of triangle $B D F$.

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-03.jpg?height=481&width=547&top_left_y=616&top_left_x=826)
Solution

Denote the lengths of $B D, B F$ and $D E$ by $x, z$ and $3 y$, respectively. Given the proportions, it follows that $E F=3 z, C E=5 y$ and $A D=x$. Therefore,

$$
\begin{aligned}
& \frac{A_{B D F}}{A_{B D E}}=\frac{z}{3 z+z} \\
& \frac{A_{B D E}}{A_{C D B}}=\frac{3 y}{3 y+5 y} \\
& \frac{A_{C D B}}{A_{A B C}}=\frac{x}{x+x}
\end{aligned}
$$

Multiplying these equations, we have

$$
\frac{A_{B D F}}{A_{A B C}}=\frac{1}{4} \cdot \frac{3}{8} \cdot \frac{1}{2}
$$

Thus,

$$
A_{B D F}=\frac{192 \cdot 3}{64}=9
$$

Note: We are denoting the area of triangle $X Y Z$ by $A_{X Y Z}$.