Right triangle in a square...

Question

On side $BC$ of the triangle $ABCD$ is point $E$, and on the side $CD$ is point $F$. If $AF=10cm$, $AE=8cm$ and $EF=6cm$ calculate a side of the square. The farthest I got is that $AEF$ is a right triangle. How do I continue?

Answer 1

Let $\angle EFC =\theta$. Then it follows that $\angle AEB =\theta$ as well. Further, say the side of the square is $x=AB=BC=CD=AD$.

From $\triangle AEB$, we have$$x=8\sin \theta \\ BE = 8\cos \theta$$

From $\triangle EFC$, $$EC=6\sin \theta$$

$$\therefore BE +EC = x = 8\cos \theta + 6\sin \theta \\ \implies 8\cos \theta +6\sin \theta = 8\sin \theta \\ \implies \sin \theta=4\cos \theta \\ \implies \frac{\sin \theta}{\sqrt{1-\sin^2 \theta}}=4 \\ \implies \sin \theta =\frac{4}{\sqrt{17}}$$

and so $$\boxed{x=\frac{32}{\sqrt{17}}}$$