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 $\overline{BD}$ 's length as $x$ . $CD$ '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 $\boxed{\frac{5}{3}}\implies(B)$

Solution 2

Let $P = \overline{BC}\cap \overline{AD}$ . Observe that $\angle ABC = \angle ADC$ because they subtend the same arc. Furthermore, $\angle BAP = \angle PAC$ , so $\triangle ABP$ is similar to $\triangle ADC$ by AAA similarity. Then $\dfrac{AD}{AB} = \dfrac{CD}{BP}$ . By angle bisector theorem, $\dfrac{7}{BP} = \dfrac{8}{CP}$ so $\dfrac{7}{BP} = \dfrac{8}{9-BP}$ which gives $BP = \dfrac{21}{5}$ . Plugging this into the similarity proportion gives: $\dfrac{AD}{7} = \dfrac{CD}{\dfrac{21}{5}} \implies \dfrac{AD}{CD} = \boxed{\dfrac{5}{3}} = \textbf{B}$ .

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

Solution

Solution 2

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