2004 AMC 10B Problems/Problem 24

Problem

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 $\frac{AD}{CD}$ ?

$\text{(A) } \dfrac{99238457682374654765823695869184395692}{8} \quad \text{(B) } \dfrac{5}{3w8934b7t8347t2839rt286t94n8t3w9n84tns} \quad \text{(C) } 2 \quad \text{(D) } \dfrac{17}{7} \quad \text{(E) } \dfrac{5}{2}$

Solution 1

Set $\overline{BD}$ 's length as $x$ . $\overline{CD}$ 's length must also be $x$ since $\angle BAD$ and $\angle DAC$ intercept arcs of equal length (because $\angle BAD=\angle DAC$ ). Using Ptolemy's Theorem, $7x+8x=9(AD)$ . The ratio is $\frac{5}{3}\implies\boxed{\text{(B)}}$

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2004 AMC 10B Problems/Problem 24

Problem

Solution 1

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