jmc

Let the radius of the circle be $r$. Then segment $AC$ has length $2r$. Recall that an inscribed angle is half the measure of the arc it cuts. Because $AC$ is a diameter of the circle, arcs $ADC$ and $ABC$ both have measure 180 degrees. Thus, angles $D$ and $B$ have measure half that, or 90 degrees. Thus, they are both right angles. Now we know that triangle $ADC$ is a 30-60-90 right triangle and that triangle $ABC$ is a 45-45-90 right triangle.

We can use the ratios of the sides in these special triangles to determine that $$\begin{array}{rl}CD& =\frac{AC}{2}=\frac{2r}{2}=r\\ AD& =DC\sqrt{3}=r\sqrt{3}\\ AB& =\frac{AC}{\sqrt{2}}=\frac{2r}{\sqrt{2}}=r\sqrt{2}\\ BC& =AB=r\sqrt{2}.\end{array}$$Now we can find the areas of triangles $ADC$ and $ABC$. $$\begin{array}{rl}{A}_{ADC}& =\frac{1}{2}(r)(r\sqrt{3})=\frac{r^2\sqrt{3}}{2}\\ A_{ABC}& =\frac{1}{2}(r\sqrt{2})(r\sqrt{2})=\frac{1}{2}(2r^2)=r^2.\end{array}$$Thus, the area of quadrilateral $ABCD$ is the sum of the areas of triangles $ADC$ and $ABC$. $$A_{ABCD}=\frac{r^2\sqrt{3}}{2}+r^2=r^2\left(\frac{\sqrt{3}}{2}+1\right)=r^2\left(\frac{\sqrt{3}+2}{2}\right).$$The area of the circle is $\pi r^2$. Thus, the ratio of the area of $ABCD$ to the area of the circle is $$\frac{r^2\left(\frac{\sqrt{3}+2}{2}\right)}{\pi r^2}=\frac{\cancel{r^2}\left(\frac{\sqrt{3}+2}{2}\right)}{\pi \cancel{r^2}}=\frac{\sqrt{3}+2}{2\pi }.$$Thus, $a=2$, $b=3$, and $c=2$. Finally, we find $a+b+c=2+3+2=\boxed{7}$.