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!

Use the well known formula of angle bisector at $A$ and equate it to projecton of rhombus sides on bisector. ( A result obtined by similar triangles)
$$ 2 b c/(b+c)  \cos A/2 = 2 s \cos A/2 $$
 $$ b c /(b+c) = s $$
 $$ s = \dfrac{72}{18} =4 $$
General formula is
$$ \frac{1}{s}= \frac{1}{b}+ \frac{1}{c} $$
NOTE: No need to specify  length of $BC=a\,!$
