TITLE: Prove that $\angle AEF =90^\circ$ given a square $ABCD$
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: $\tan(\angle AED) = 2.$ 
$\tan(\angle FEC) = (1/2) = $cot$(\angle AED).$
Further, $\angle AED$ and $\angle FEC$ are both acute.
Therefore, $\angle AED + \angle FEC = 90^\circ.$