A rhombus is inscribed in triangle ABC > A rhombus is inscribed in triangle $ABC$ with $A$ as one of its vertices. Two sides of the rhombus lie along $AB$ and $AC$, with $AC=6, AB = 12$, and $BC = 8$. Find the length of a side of the rhombus. So far by using heron's formula I got that the area of $ABC$ is $\sqrt {455}$, and letting $s$ be the side length of the rhombus, that $s<6$ (naturally). Also, letting the vertice of rhombus on $AB$ be $P$ and the vertice of rhombus on $AC$ be $Q$, $BP=12-s$ and $QC = 6-s$. However, I'm not sure how to carry on from here. Any help would be appreciated!

Let $R$ be a vertex of the rhombus on the side $BC$.

Then, since $PR$ is parallel to $AC$, we see that $\triangle{BPR}$ and $\triangle{BAC}$ are similar.

So, we have $$BP:BA=PR:AC\implies 12-s:12=s:6\implies s=4$$