jmc

Suppose that $AD$ and $BC$ intersect at $E$.



Since $\angle ADC$ and $\angle ABC$ cut the same arc of the circumscribed circle, the Inscribed Angle Theorem implies that $$\angle ABC=\angle ADC.$$Also, $\angle EAB=\angle CAD$, so $△ABE$ is similar to $△ADC$, and $$\frac{AD}{CD}=\frac{AB}{BE}.$$By the Angle Bisector Theorem, $$\frac{BE}{EC}=\frac{AB}{AC},$$so $$BE=\frac{AB}{AC}\cdot EC=\frac{AB}{AC}(BC-BE)\text{and}BE=\frac{AB\cdot BC}{AB+AC}.$$Hence $$\frac{AD}{CD}=\frac{AB}{BE}=\frac{AB+AC}{BC}=\frac{7+8}{9}=\boxed{\frac{5}{3}}.$$