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 $\frac{AD}{CD}$ ?

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

Solution 1 - (Ptolemy's Theorem)

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 - Similarity Proportion

$[asy] import graph; import geometry; import markers; unitsize(0.5 cm); pair A, B, C, D, E, I; A = (11/3,8*sqrt(5)/3); B = (0,0); C = (9,0); I = incenter(A,B,C); D = intersectionpoint(I--(I + 2*(I - A)), circumcircle(A,B,C)); E = extension(A,D,B,C); draw(A--B--C--cycle); draw(circumcircle(A,B,C)); draw(D--A); draw(D--B); draw(D--C); label("$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 $\angle ABC \cong \angle ADC$ because they both subtend arc $\overarc{AC}.$

Furthermore, $\angle BAE \cong \angle EAC$ because $\overline{AE}$ is an angle bisector, so $\triangle ABE \sim \triangle ADC$ by $\text{AA}$ similarity. Then $\dfrac{AD}{AB} = \dfrac{CD}{BE}$ . By the Angle Bisector Theorem, $\dfrac{7}{BE} = \dfrac{8}{CE}$ , so $\dfrac{7}{BE} = \dfrac{8}{9-BE}$ . This in turn gives $BE = \frac{21}{5}$ . Plugging this into the similarity proportion gives: $\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}$ .

Solution 3 (Angle Bisector Theorem)

Similar to solution 1, let $E$ be the intersection of diagonals $AC and$ BD $. We know that$ \overline{AD} $bisects$ \angle BAC $, so$ \angle BAD = \angle CAD $. Additionally,$ \angle BAD $and$ \angle BCD $subtend the same arc, giving$ \angle BAD = \angle BCD $. Similarly,$ \angle CAD = \angle CBD $and$ \angle ABC = \angle ADC$.

These angle relationships tell us that$ (Error compiling LaTeX. Unknown error_msg)\triangle ABE\sim \triangle ADC $by AA Similarity, so$ AD/CD = AB/BE $. By the angle bisector theorem,$ AB/BE = AC/CE$. Hence, <cmath>\frac{AB}{BE} = \frac{AC}{CE} = \frac{AB + AC}{BE + CE} = \frac{AB + AC}{BC} = \frac{7 + 8}{9} = \frac{15}{9} = \boxed{\frac{5}{3}}.</cmath>

~vaporwave

== Solution 4 (Ptolemy's Theorem)

Set$ (Error compiling LaTeX. Unknown error_msg)\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)$

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

Contents

Problem

Solution 1 - (Ptolemy's Theorem)

Solution 2 - Similarity Proportion

Solution 3 (Angle Bisector Theorem)

See Also