Find the side of an equilateral triangle inscribed in a circle. 
Excuse the poor drawing. 

$\triangle CDE$ is an equilateral triangle inscribed inside a circle, with side length $16$. Let $F$ be the midpoint of $DE$. Points $G$ and $H$ are on the circle so that $\triangle FGH$ is an equilateral. Find the side length of $\triangle FGH$ and express it as $a\sqrt{b}-c$ where $a, b, c$ are positive integers. 

My attempt
$$\theta \text{ in the Large } \triangle = \theta \text{ in the Small } \triangle = 60^{\circ}$$
$$(\text{side of the Large } \triangle) \cdot \sin 60^{\circ} + (\text{side of the Small } \triangle )\cdot \sin 60^{\circ} = 2\cdot(\text{radius of circle})$$
Solving for the small side, I got $\dfrac{64}{3}-16$, which doesn't match the required form. 
Thanks for any help.
 A: This answer may not be the simplest, but it is straightforward.

I particularized the problem by making one side of the large equilateral triangle the segment between the points $(-8,0)$ and $(8,0)$. Then simple geometry tells us the third vertex is at $(8\sqrt{3},0)$, the circumcenter of the triangle is at $A(0,\dfrac{8\sqrt{3}}3)$, and the radius of the circumcircle is $\dfrac{16\sqrt{3}}3$.
The equation of the circumcircle is then
$$x^2+\left(y-\frac{8\sqrt{3}}3\right)^2=\left(\frac{16\sqrt{3}}3\right)^2$$
$$x^2+\left(y-\frac{8\sqrt{3}}3\right)^2=\frac{256}3$$
The one side of the small equilateral triangle, $\overline{FH}$, is on the line $y=-\sqrt{3}x$. That gives us two equations in two unknowns for the coordinates of point $H$ which is on both the circle and the line. Substitute the expression for $y$ in the linear equation into the quadratic equation and we get a quadratic equation for $x$:
$$x^2+\left(-\sqrt 3x-\frac{8\sqrt 3}3\right)^2=\frac{256}3$$
$$4x^2+16x-64=0$$
$$x^2+4x-16=0$$
Solving this gives us this positive value for $x$:
$$x=2\sqrt 5-2$$
The side of the small equilateral triangle is twice the $x$-coordinate of point $H$, so the side of the triangle is
$$4\sqrt 5-4$$
The final answer to your problem, given the side is $a\sqrt b-c$, is

$$a=4, \quad b=5, \quad c=4$$

I checked this answer numerically with Geogebra, the source of my diagram above, and my answer checks.
A: Let $R$ be the radius of circle.  Draw a perpendicular to any of sides of large equilateral $\triangle CDE$ to obtain a right triangle by which we can calculate the side length $L$ of large equilateral triangle as follows $$L=2R\cos 30^\circ=16$$  $$\implies 2R\frac{\sqrt{3}}{2}=16 \implies \color{blue}{R=\frac{16}{\sqrt{3}}}$$ Let $l$ be side length of small equilateral $\triangle FGH$ & draw a perpendicular say $AM$, from center $A$ passing through $F$, to the opposite side $GH$ of small $\triangle FGH$ Then we have $$AM=AF+FM=R\sin 30^\circ+l\sin 60^\circ=\frac{16}{\sqrt{3}}\frac{1}{2}+l\frac{\sqrt{3}}{2}=\color{red}{\frac{16+3l}{2\sqrt{3}}}$$ in right $\triangle AMG$ applying pythagorean theorem as follows $$AG^2=AM^2+GM^2$$ $$\left(\frac{16}{\sqrt{3}}\right)^2=\left(\frac{16+3l}{2\sqrt{3}}\right)^2+\left(\frac{l}{2}\right)^2$$  $$ 12l^2+96l-768=0$$ Solving for $l$ as follows $$l=\frac{-96\pm\sqrt{(96)^2-4(12)(-768)}}{2(12)}$$ $$l=\frac{-96\pm96\sqrt{5}}{24}=-4\pm 4\sqrt{5}$$ but $l>0$ hence the side of the small equilateral $\triangle FGH$ $$l=\color{red}{4\sqrt{5}-4}$$ Now, comparing the side $l$ with $\color{red}{a\sqrt{b}-c}$ we get $$\color{blue}{a=4, b=5, c=4}$$
A: Extending $\overline{FG}$ crosses $\overline{CE}$ at it midpoint, $F^\prime$ and hits the circle again at $G^\prime$. Clearly, $\overline{F^\prime G^\prime}$ is the side of a triangle congruent to $\triangle FGH$.

Define $p := |\overline{DF}| = |\overline{EF}| = |\overline{FF^\prime}|$ (where the last equality follows from the Triangle Midsegment Theorem) and $q := |\overline{FG}| = |\overline{F^\prime G^\prime}|$. By the Power of a Point Theorem applied to point $F$, we have
$$|\overline{DF}|\;|\overline{EF}| = |\overline{GF}|\;|\overline{G^\prime F}|$$
$$\to \quad p^2 = q ( p + q )
\quad \to \quad q^2 + p q - p^2 = 0
\quad \to \quad q = \frac{p}{2}\left( \sqrt{5} - 1 \right) \tag{$\star$}$$
(Note that we have taken the positive root of the quadratic equation.)
Then, since $p=8$,

$$q = 4(\sqrt{5}-1) = 4\sqrt{5}-4 \qquad\to\qquad (a,b,c) = (4,5,4)$$

