Find side of an equilateral triangle inscribed in a rhombus The lengths of the diagonals of a rhombus are 6 and 8. An equilateral triangle inscribed in this
rhombus has one vertex at an end-point of the shorter diagonal and one side parallel to the longer
diagonal. Determine the length of a side of this triangle. 
Express your answer in the form $k\left(4 \sqrt{3} − 3\right)$ where k is a vulgar fraction.


Hopefully one can see anything in this drawing...
You know that all the green and all the blue lines are equal. Additionally, you know that the top green line is parallel to one of the diagonals and that the diagonals are perpendicular. Therefore, the top green line is perpendicular to the short diagonal as well.
From the Pythagorean theorem follows
$$b^2-\left(\frac{b}{2}\right)^2=\left(n-x\right)^2$$
where $n$ is the short diagonal, $a$ is the length we are looking for and $x$ is the orange bit. This equation still has an $x$ in it; we need it in terms of $n$ and $m$.
Using the Intercept theorem, one can infer
$$\frac{x}{\frac{b}{2}}=\frac{\frac{n}{2}}{\frac{m}{2}} \implies x=\frac{nb}{2m}$$
where $m$ denotes the long diagonal.
Algebra:
\begin{align*}\frac{3}{4}b^2=\left(n-x\right)^2&=\left(n-\frac{nb}{2m}\right)^2\\
\frac{\sqrt{3}}{2}b&=n-\frac{nb}{2m}\\
\frac{\sqrt{3}}{2}b+\frac{nb}{2m}&=n\\
b\left(\frac{\sqrt{3}}{2}+\frac{n}{2m}\right)&=n\\
b=\frac{n}{\frac{\sqrt{3}}{2}+\frac{n}{2m}}&=\frac{2mn}{\sqrt{3}m+n}\end{align*}
Substituting the given values:
$$b=\frac{96}{8\sqrt{3}+6}=\frac{48}{4\sqrt{3}+3}=\frac{48\left(4\sqrt{3}-3\right)}{\left(4\sqrt{3}+3\right)\left(4\sqrt{3}-3\right)}=\frac{48\left(4\sqrt{3}-3\right)}{\left(4\sqrt{3}\right)^2-3^2}=\frac{16}{13}\left(4\sqrt{3}-3\right)$$
