In the figure below, triangle $A B C$ is a right triangle at $C$ and both $B C D E$ and $C A F G$ are squares. If the product of the areas of triangles $E A B$ and $B F A$ is 64, determine the area of triangle $A B C$.

![](https://cdn.mathpix.com/cropped/2024_05_01_10107d2fba0ec26bc845g-17.jpg?height=649&width=985&top_left_y=555&top_left_x=638)
Solution

Since $D A$ is parallel to $E B$, the area of triangle $A E B$ is $\frac{E B \cdot B C}{2}=\frac{B C^{2}}{2}$. Similarly, the area of triangle $A B F$ is $\frac{A C^{2}}{2}$. Therefore,

$$
\begin{aligned}
{[A B E] \cdot[A B F] } & =\frac{B C^{2}}{2} \cdot \frac{A C^{2}}{2} \\
64 \cdot 4 & =B C^{2} \cdot A C^{2} \\
16 & =B C \cdot A C
\end{aligned}
$$

Thus, the area of triangle $A B C$ is $\frac{B C \cdot A C}{2}=\frac{16}{2}=8$.