smc

As in the diagram, suppose the rhombus $ADEF$ is inscribed in $△ABC$ with $D$ on $\overline{AC}$, $E$ on $\overline{BC}$, and $F$ on $\overline{AB}$. Let the side length of the rhombus be $x$. Because a rhombus is a parallelogram, its opposite sides are parallel. Thus, by AA similarity, $△BFE\sim △EDC$, and so $\frac{FE}{FB}=\frac{DC}{DE}$. Because $\overline{FE}$ and $\overline{DE}$ are sides of the rhombus, they both have length $x$. Furthermore, $FB=12-x$, and $DC=6-x$, because, combined with sides of the rhombus, they form sides of the triangle. Thus, by substituting into the proportion derived above, we see that: $$\begin{array}{rl}\frac{FE}{FB}& =\frac{DC}{DE}\\ \frac{x}{12-x}& =\frac{6-x}{x}\\ x^2& =72-12x-6x+x^2\\ 18x& =72\\ x& =4\end{array}$$ Thus, the side of the rhombus has length $\boxed{4}$.