Prove that $\angle AEF =90^\circ$ given a square $ABCD$

Triangles EDA FCE are similar. Scale factor =2.

$$ FC= EC/2,\;DE= AD/2 \;$$

Since the triangles are similar corresponding angles are same. (Given square contains a right angle).

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: $$AE^2+EF^2=\frac{25}{16}a^2=AF^2$$

$\tan(\angle AED) = 2.$
$\tan(\angle FEC) = (1/2) = $cot$(\angle AED).$

Further, $\angle AED$ and $\angle FEC$ are both acute.

Therefore, $\angle AED + \angle FEC = 90^\circ.$