In $\triangle ABC$ we have $AB=7$, $AC=8$, and $BC=9$. Point $D$ is on the circumscribed circle of the triangle so that $\overline{AD}$ bisects $\angle BAC$. What is the value of $AD/CD$?
Suppose that $AD$ and $BC$ intersect at $E$.

[asy]
pair A,B,C,D,I;
A=(-9,-4.36);
B=(-7,7.14);
C=(8,-6);
D=(7.5,6.61);
I=(2.7,3);
draw(Circle((0,0),10));
draw(A--B--C--cycle,linewidth(0.7));
draw(B--D--C);
draw(A--D);
label("$E$",I,S);
label("$B$",B,N);
label("$D$",D,NE);
label("$C$",C,E);
label("$A$",A,SW);
[/asy]

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 $\triangle ABE$ is similar to $\triangle ADC$, and \[
\frac{AD}{CD} = \frac{AB}{BE}.
\]By the Angle Bisector Theorem, \[
\frac{BE}{EC} = \frac{AB}{AC},
\]so \[
BE = \frac{AB}{AC} \cdot EC = \frac{AB}{AC}(BC - BE)
\quad\text{and}\quad BE = \frac{AB\cdot BC}{AB+AC}.
\]Hence \[
\frac{AD}{CD} = \frac{AB}{BE} = \frac{AB+AC}{BC} =
\frac{7+8}{9} = \boxed{\frac{5}{3}}.
\]