Rinse and repeat ....

First, using the Cosine Law we have that

$(BD)^{2} = (AB)^{2} + (AD)^{2} - 2*(AB)(AD)\cos(\angle BAD) =$

$2 - 2*(-\frac{1}{3}) = \frac{8}{3} \Longrightarrow BD = \sqrt{\frac{8}{3}}.$

Now let $CD = \cos(\angle ABC) = x$ . Since $ABCD$ is cyclic we have that

$\cos(\angle ADC) = -\cos(\angle ABC) = -x$ and

$\cos(\angle BCD) = -\cos(\angle BAD) = \frac{1}{3}$ .

Nwxt, since $\Delta ABC$ and $\Delta ADC$ have chord $AC$ in common, we can use the Cosine Law to set up the following equation:

$1^{2} + x^{2} - 2(1)(x)\cos(\angle ADC) = 1^{2} + (BC)^{2} - 2(1)(BC)\cos(\angle ABC)$

$\Longrightarrow 1 + x^{2} - 2x*(-x) = 1 + (BC)^{2} - 2(BC)(x)$

$\Longrightarrow (BC)^{2} - 2x*(BC) - 3x^{2} = 0 \Longrightarrow (BC - 3x)(BC + x) = 0.$

Now since $BC \gt 0$ we have that $BC = 3x$ . But since the sides $DC$ and $BC$ are in the ratio $x : 3x \Longrightarrow 1 : 3$ and $\cos(\angle BCD) = \frac{1}{3}$ we can conclude that $\Delta BCD$ is a right triangle with $\angle BDC = 90^{\circ}$ . This implies that $BC$ is a diameter of the circle in which $ABCD$ is inscribed.

Using Pythagoras, we then have that

$(BC)^{2} = (CD)^{2} + (BD)^{2} \Longrightarrow 9x^{2} = x^{2} + \frac{8}{3} \Longrightarrow x = CD = \dfrac{\sqrt{3}}{3},$

as $x \gt 0$ , and so $BC = 3x = \sqrt{3}.$

So with these side lengths, we can apply the formula for the area of a cyclic quadrilateral, namely

$\sqrt{(s - t)(s - u)(s - v)(s - w)}$

where $t,u,v,w$ are the side lengths and $s = \dfrac{t + u + v + w}{2}.$

With $s = \dfrac{1 + 1 + \frac{\sqrt{3}}{3} + \sqrt{3}}{2} = 1 + \dfrac{2}{3}\sqrt{3}$ the area calculation becomes

$\sqrt{(\dfrac{2\sqrt{3}}{3})^{2}(1 + \dfrac{\sqrt{3}}{3})(1 - \dfrac{\sqrt{3}}{3})} = \dfrac{2}{3}\sqrt{2}.$

The area of the circle is $\pi*(\dfrac{\sqrt{3}}{2})^{2} = \dfrac{3}{4}\pi$ ,

so the desired ratio is $\dfrac{\frac{2}{3}\sqrt{2}}{\frac{3}{4}\pi} = \dfrac{8\sqrt{2}}{9\pi}$ .

Thus $a + b + c = 8 + 2 + 9 = \boxed{19}$ .