In the following drawing, $ABCD$ is a square and points $E$ and $F$ are on sides $BC$ and $CD$ such that $AEF$ is a right triangle, $AE=4$ and $EF=3$. What is the area of the square?

![](https://cdn.mathpix.com/cropped/2024_05_01_10107d2fba0ec26bc845g-10.jpg?height=417&width=437&top_left_y=2276&top_left_x=707)
Solution

If $\angle B A E=\alpha$, since $\angle A B C=\angle A E F=90^{\circ}$, we have $\angle B E A=90^{\circ}-\alpha$ and $\angle F E C=\alpha$. Consequently, triangles $B E A$ and $C F E$ are similar and

$$
\frac{B E}{C F}=\frac{A B}{E C}=\frac{A E}{E F}=\frac{4}{3}
$$

If $B E=x$, then $C F=\frac{3 x}{4}$ and $\frac{A B}{A B-x}=\frac{4}{3}$, which means $A B=4 x$. Applying the Pythagorean Theorem to triangles $E A F$ and $A F D$, we have

$$
\begin{aligned}
A E^{2}+E F^{2} & =A F^{2} \\
3^{2}+4^{2} & =A D^{2}+F D^{2} \\
25 & =A B^{2}+(A B-C F)^{2} \\
25 & =16 x^{2}+\frac{169 x^{2}}{16} \\
25 & =\frac{425 x^{2}}{16} \\
x^{2} & =\frac{16}{17}
\end{aligned}
$$

Therefore, the area of the square is $A B^{2}=16 x^{2}=\frac{256}{17}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_10107d2fba0ec26bc845g-11.jpg?height=420&width=434&top_left_y=2000&top_left_x=911)