2004 AMC 10B Problems/Problem 24

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In triangle $ABC$ we have $AB=7$, $AC=8$, $BC=9$. Point $D$ is on the circumscribed circle of the triangle so that $AD$ bisects angle $BAC$. What is the value of $AD/CD$? \\ $A. \dfrac{9}{8}$ $B. \dfrac{5}{3}$ $C. 2$ $D. \dfrac{17}{7}$ $E. \dfrac{5}{2}$

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