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.

Ignoring the similar triangles for variety...
Let the side length of the square be $s$. We have $|AF|^2 = |AB|^2 +|BF|^2 = s^2 + \left(\frac 34 s\right)^2 = \frac{25}{16}s^2$
Then $|AE|^2 = |AD|^2 +|DE|^2 = s^2 + \left(\frac 12 s\right)^2  =  \frac{5}{4}s^2 =  \frac{20}{16}s^2$
and  $|EF|^2 = |EB|^2 +|BF|^2 = \left(\frac 12 s\right)^2 + \left(\frac 14 s\right)^2  =  \frac{5}{16}s^2$
giving $|AF|^2 = |AE|^2 +|EF|^2$ as required
