Prove that $\angle AEF =90^\circ$ given a square $ABCD$ Let $ABCD$ be a square and $E\in CD$ such that $DE=EC$. Let $F\in BC$ such that $BC=4FC$. Prove that $\angle AEF =90^\circ$.

My attempt:
Proving that $\angle AEF =90^\circ$ is the same as proving that $\triangle AEF$ is a right triangle. In other words, we wish to prove that $AE^2 +EF^2 = AF^2$. I’ve tried a lot of methods to reach this point but none of them worked.

The idea in your attempt definitely works here (although it is not the shortest proof, see the answer by Narasimham): Let $a$ be the sidelength of the square. Then we have \begin{align*}
AE^2&=AD^2+DE^2=a^2+\left(\frac{1}{2}a\right)^2\\
EF^2&=EC^2+CF^2=\left(\frac{1}{2}a\right)^2+\left(\frac{1}{4}a\right)^2\\
AF^2&=AB^2+BF^2=a^2+\left(\frac{3}{4}a\right)^2
\end{align*}
And thus: $$AE^2+EF^2=\frac{25}{16}a^2=AF^2$$
