In triangle $ABC$, $AB=9,AC=10,BC=8$. Point $D$ is on the circumscribed circle so that $AD$ bisects $\angle BAC$. What is $\dfrac{AD}{CD}$?

1. Since $AD$ bisects $\angle BAC$, the chords $BD$ and $CD$ are equal. Ptolemy's theorem (45) on the cyclic quadrilateral $ABDC$ gives $AD\cdot BC=AB\cdot CD+AC\cdot BD=CD\,(AB+AC)$, so $\dfrac{AD}{CD}=\dfrac{AB+AC}{BC}$.
2. With $AB=9,AC=10,BC=8$: $\dfrac{AD}{CD}=\dfrac{19}8$.
3. The answer is $\boxed{\dfrac{19}8}$.