2004 AMC 10B Problem 24

Let $\overline{AD}$ meet $\overline{BC}$ at $E.$ Since $\angle ABC$ and $\angle ADC$ subtend the same arc, they are equal, and $\angle EAB = \angle CAD,$ so $\triangle ABE \sim \triangle ADC.$ Hence $\dfrac{AD}{CD} = \dfrac{AB}{BE}.$ By the Angle Bisector Theorem, $\dfrac{BE}{EC} = \dfrac{AB}{AC},$ so $BE = \dfrac{AB \cdot BC}{AB + AC} = \dfrac{7 \cdot 9}{15}.$ Therefore $$\begin{aligned} \dfrac{AD}{CD} &= \dfrac{AB}{BE} = \dfrac{AB + AC}{BC} \\ &= \dfrac{15}{9} = \dfrac{5}{3}. \end{aligned}$$ Thus, the correct answer is B.