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Difference between revisions of "2004 AMC 10B Problems/Problem 24"
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<math>\text{(A) } \dfrac{9}{8} \quad \text{(B) } \dfrac{5}{3} \quad \text{(C) } 2 \quad \text{(D) } \dfrac{17}{7} \quad \text{(E) } \dfrac{5}{2}</math>
 
<math>\text{(A) } \dfrac{9}{8} \quad \text{(B) } \dfrac{5}{3} \quad \text{(C) } 2 \quad \text{(D) } \dfrac{17}{7} \quad \text{(E) } \dfrac{5}{2}</math>
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== Solution ==
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Set <math>BD</math>'s length as <math>x</math>. <math>CD</math>'s length must also be <math>x</math>. Using Ptolemy's Theorem, <math>7x+8x=9(AD)</math>. The ratio is <math> \boxed{\frac{5}{3}}\implies(C)</math>

Revision as of 11:34, 1 May 2011

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 $AD/CD$?

$\text{(A) } \dfrac{9}{8} \quad \text{(B) } \dfrac{5}{3} \quad \text{(C) } 2 \quad \text{(D) } \dfrac{17}{7} \quad \text{(E) } \dfrac{5}{2}$


Solution

Set $BD$'s length as $x$. $CD$'s length must also be $x$. Using Ptolemy's Theorem, $7x+8x=9(AD)$. The ratio is $\boxed{\frac{5}{3}}\implies(C)$

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