Inscribed Right Triangle

(Edit 6/18/23: The image has been lost, but I’ll leave the text in case I ever have the chance to reconstruct it.) Here’s a fun puzzle (via Brilliant.org): What is the area of the square $ABCD$?

There may be a simpler approach; my solution wound up being more complicated than I expected.

Since $\Delta AEF$ is a right triangle, $AE = 5$ and $\sin{\gamma} = 3/5$.

Let $(FD, BE, AD) = (a, b, c)$. We can disregard the other segments along the perimeter. We know that the area of the $ABCD = c^2$, so our goal is to calculate $c$.

For my next step, I rearranged the triangles. Since $AB = AD$, we can move $\Delta AEB$ to connect $\overline{AD}$ to $\overline{AB}$ to create $\Delta AFE$. What do we know about this triangle?

$AF = 4$ and $AE = 5$. $m\angle{FAE} = \beta + \delta = \eta – \gamma$ (where $\eta = 90^\circ = \pi/2$). Using the Law of Cosines, we can calculate $a + b  = BE + DF = EF$. Specifically, $EF^2 = 4^2 + 5^2 – 2(4)(5)\cos{(\eta – \gamma)}$. Using an identity, $\cos{(\eta – \gamma)} = \sin{\gamma}$, so $EF^2 = 16 + 25 – 40\sin{\gamma} = 41 – 40(3/5) = 17$ and $EF = \sqrt{17}$.

Using Heron’s formula, the area of $\Delta AFE = \sqrt{(5 + 4 + \sqrt{17})(5 + 4 – \sqrt{17})(5 – 4 + \sqrt{17})(-5 + 4 + \sqrt{17})}/4$ $= \sqrt{(9 + \sqrt{17})(9 – \sqrt{17})(\sqrt{17} + 1)(\sqrt{17} – 1)}/4$ $= \sqrt{(81 – 17)(17 – 1)}/4 = \sqrt{1024}/4 = 8$.

That means the altitude of $\Delta AFE = 2 \times 8 / \sqrt{17} = 16 / \sqrt{17} = AD$, and the area of the square is $(16/\sqrt{17})^2 = \frac{256}{17}$.