# 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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• Reduce fractions to lowest terms and enter in the form 7/9.
• Numbers involving pi should be written as 7pi or 7pi/3 as appropriate.
• Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate.
• Exponents should be entered in the form 10^10.
• If the problem is multiple choice, enter the appropriate (capital) letter.
• Enter points with parentheses, like so: (4,5)
• Complex numbers should be entered in rectangular form unless otherwise specified, like so: 3+4i. If there is no real component, enter only the imaginary component (i.e. 2i, NOT 0+2i).