Right $\triangle ABC$ has $AB=5, BC=12,$ and $AC=13.$ Square $XYZW$ is inscribed in $\triangle ABC$ with $X$ and $Y$ on $\overline{AC}, W$ on $\overline{AB},$ and $Z$ on $\overline{BC}.$ What is the side length of the square?

Let the side of the square be $x$; the right angle is at $B$. By $\triangle CYZ\sim \triangle CBA$ (right angles at $Y$ and $B$, common $\angle C$): $CZ:13=x:5$, so $CZ=\frac{13x}5$.
Since $WZ\parallel AC$, $\triangle WBZ\sim\triangle ABC$: $ZB:12=x:13$, so $BZ=\frac{12x}{13}$.
$CZ+BZ=BC=12$: $\frac{13x}5+\frac{12x}{13}=12\Rightarrow \frac{169x+60x}{65}=12\Rightarrow 229x=780\Rightarrow x=\boxed{\frac{780}{229}}$.