<s>[INST] 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.
 [/INST] 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: $$AE^2+EF^2=\frac{25}{16}a^2=AF^2$$
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