What is the length of a side of square $DCEF$ where $AB=3, BC=4, AC=5$ in the following diagram? 

$DCEF$ is a square and $\triangle ABC$ is an internal triangle. $AB=3,$
$BC=4, AC=5$. What is the length of the square?

I tried multiple times and ended up by converting variables again and again. Can someone give me a hint to solve this?

The key is $$3^2 + 4^2 = 5^2 \quad\implies\quad \angle ABC = 90^\circ$$ 
As a consequence,
$$\angle ABD = \angle BCF\quad\implies\quad \triangle ABD \simeq \triangle BCF$$
Let $s = CF$ be the side of the square, we have
$$BD : AB = CF : BC \iff BD = \frac{3s}{4} \implies BF = DF - BD = \frac{s}{4}$$
Apply Pythagoras theorem to $\triangle BCF$, we get
$$BF^2 + FC^2 = 4^2\quad\iff\quad\frac{s^2}{16} + s^2 = 4^2\quad\implies\quad s = \frac{16}{\sqrt{17}}$$
