2004 AMC 10B Problems/Problem 24

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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2004 AMC 10B Problems/Problem 24

Solution