# Problem #1137

 1137 In $\bigtriangleup ABC$ we have $AB = 7$, $AC = 8$, and $BC = 9$. Point $D$ is on the circumscribed circle of the triangle so that $AD$ bisects $\angle BAC$. What is the value of $\frac{AD}{CD}$? $\mathrm{(A) \ } \frac{9}{8} \qquad \mathrm{(B) \ } \frac{5}{3} \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ } \frac{17}{7} \qquad \mathrm{(E) \ } \frac{5}{2}$ This problem is copyrighted by the American Mathematics Competitions.
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