# Difference between revisions of "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 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 $\frac{5}{3}\implies\boxed{\text{(B)}}$

## Solution 2

$[asy] import graph; import geometry; import markers; unitsize(0.5 cm); pair A, B, C, D, E, I; A = (11/3,8*sqrt(5)/3); B = (0,0); C = (9,0); I = incenter(A,B,C); D = intersectionpoint(I--(I + 2*(I - A)), circumcircle(A,B,C)); E = extension(A,D,B,C); draw(A--B--C--cycle); draw(circumcircle(A,B,C)); draw(D--A); draw(D--B); draw(D--C); label("A", A, N); label("B", B, SW); label("C", C, SE); label("D", D, S); label("E", E, NE); markangle(radius = 20,B, A, C, marker(markinterval(2,stickframe(1,2mm),true))); markangle(radius = 20,B, C, D, marker(markinterval(1,stickframe(1,2mm),true))); markangle(radius = 20,D, B, C, marker(markinterval(1,stickframe(1,2mm),true))); markangle(radius = 20,C, B, A, marker(markinterval(1,stickframe(2,2mm),true))); markangle(radius = 20,C, D, A, marker(markinterval(1,stickframe(2,2mm),true))); [/asy]$ Let $E = \overline{BC}\cap \overline{AD}$. Observe that $\angle ABC = \angle ADC$ because they subtend the same arc. Furthermore, $\angle BAE = \angle EAC$, so $\triangle ABE$ is similar to $\triangle ADC$ by AAA similarity. Then $\dfrac{AD}{AB} = \dfrac{CD}{BE}$. By angle bisector theorem, $\dfrac{7}{BE} = \dfrac{8}{CE}$ so $\dfrac{7}{BE} = \dfrac{8}{9-BE}$ which gives $BE = \dfrac{21}{5}$. Plugging this into the similarity proportion gives: $\dfrac{AD}{7} = \dfrac{CD}{\dfrac{21}{5}} \implies \dfrac{AD}{CD} = \boxed{\dfrac{5}{3}} = \textbf{B}$.