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Difference between revisions of "2004 AMC 10B Problems/Problem 24"

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In triangle <math>ABC</math> we have <math>AB=7</math>, <math>AC=8</math>, <math>BC=9</math>. Point <math>D</math> is on the circumscribed circle of the triangle so that <math>AD</math> bisects angle <math>BAC</math>. What is the value of <math>AD/CD</math>?
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==Problem==
  
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<math>\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}</math>
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In triangle <math>ABC</math> we have <math>AB=7</math>, <math>AC=8</math>, <math>BC=9</math>. Point <math>D</math> is on the circumscribed circle of the triangle so that <math>AD</math> bisects angle <math>BAC</math>. What is the value of <math>\frac{AD}{CD}</math>?
  
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<math>\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}</math>
  
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== Solution ==
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== Solution 1 - (Ptolemy's Theorem) ==
  
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Set <math>\segment BD</math>'s length as <math>x</math>. <math>CD</math>'s length must also be <math>x</math> since <math> \angle BAD </math> and <math> \angle DAC </math> intercept arcs of equal length. Using [[Ptolemy's Theorem]], <math>7x+8x=9(AD)</math>. The ratio is <math>\boxed{\frac{5}{3}}\implies(B)</math>
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Set <math>\overline{BD}</math>'s length as <math>x</math>. <math>\overline{CD}</math>'s length must also be <math>x</math> since <math>\angle BAD</math> and <math>\angle DAC</math> intercept arcs of equal length (because <math>\angle BAD=\angle DAC</math>). Using [[Ptolemy's Theorem]], <math>7x+8x=9(AD)</math>. The ratio is <math>\frac{5}{3}\implies\boxed{\text{(B)}}</math>
  
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==Solution 2==
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==Solution 2 - Similarity Proportion==
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<asy>
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import graph;
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import geometry;
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import markers;
  
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Let <math>P = \overline{BC}\cap \overline{AD}</math>.  Observe that <math>\angle ABC = \angle ADC</math> because they subtend the same arc.  Furthermore, <math>\angle BAP = \angle PAC</math>, so <math>\triangle ABP</math> is similar to <math>\triangle ADC</math> by AAA similarity. Then <math>\dfrac{AD}{AB} = \dfrac{CD}{BP}</math>.  By angle bisector theorem, <math>\dfrac{7}{BP} = \dfrac{8}{CP}</math> so <math>\dfrac{7}{BP} = \dfrac{8}{9-BP}</math> which gives <math>BP = \dfrac{21}{5}</math>.  Plugging this into the similarity proportion gives:  <math>\dfrac{AD}{7} = \dfrac{CD}{\dfrac{21}{5}} \implies \dfrac{AD}{CD} = \boxed{\dfrac{5}{3}} = \textbf{B}</math>.
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unitsize(0.5 cm);
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pair A, B, C, D, E, I;
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A = (11/3,8*sqrt(5)/3);
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B = (0,0);
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C = (9,0);
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I = incenter(A,B,C);
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D = intersectionpoint(I--(I + 2*(I - A)), circumcircle(A,B,C));
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E = extension(A,D,B,C);
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draw(A--B--C--cycle);
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draw(circumcircle(A,B,C));
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draw(D--A);
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draw(D--B);
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draw(D--C);
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label("$A$", A, N);
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label("$B$", B, SW);
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label("$C$", C, SE);
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label("$D$", D, S);
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label("$E$", E, NE);
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markangle(radius = 20,B, A, C, marker(markinterval(2,stickframe(1,2mm),true)));
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markangle(radius = 20,B, C, D, marker(markinterval(1,stickframe(1,2mm),true)));
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markangle(radius = 20,D, B, C, marker(markinterval(1,stickframe(1,2mm),true)));
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markangle(radius = 20,C, B, A, marker(markinterval(1,stickframe(2,2mm),true)));
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markangle(radius = 20,C, D, A, marker(markinterval(1,stickframe(2,2mm),true)));
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</asy>
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Let <math>E = \overline{BC}\cap \overline{AD}</math>.  Observe that <math>\angle ABC \cong \angle ADC</math> because they both subtend arc <math>\overarc{AC}.</math>  
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Furthermore, <math>\angle BAE \cong \angle EAC</math> because <math>\overline{AE}</math> is an angle bisector, so <math>\triangle ABE \sim \triangle ADC</math> by <math>\text{AA}</math> similarity. Then <math>\dfrac{AD}{AB} = \dfrac{CD}{BE}</math>.  By the [[Angle Bisector Theorem]], <math>\dfrac{7}{BE} = \dfrac{8}{CE}</math>, so <math>\dfrac{7}{BE} = \dfrac{8}{9-BE}</math>. This in turn gives <math>BE = \frac{21}{5}</math>.  Plugging this into the similarity proportion gives:   
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<math>\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}</math>.
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==Solution 3 - Angle Bisector Theorem==
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We know that <math>\overline{AD}</math> bisects <math>\angle BAC</math>, so <math>\angle BAD = \angle CAD</math>. Additionally, <math>\angle BAD</math> and <math>\angle BCD</math> subtend the same arc, giving <math>\angle BAD = \angle BCD</math>. Similarly, <math>\angle CAD = \angle CBD</math> and <math>\angle ABC = \angle ADC</math>.
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 +
These angle relationships tell us that <math>\triangle ABE\sim \triangle ADC</math> by AA Similarity, so <math>AD/CD = AB/BE</math>. By the angle bisector theorem, <math>AB/BE = AC/CE</math>. Hence,
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<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>
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(Where did E come from?????)
 +
 
 +
--vaporwave
  
 
== See Also ==
 
== See Also ==

Latest revision as of 14:41, 2 November 2024

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

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 $\triangle ABE\sim \triangle ADC$ by AA Similarity, so $AD/CD = AB/BE$. By the angle bisector theorem, $AB/BE = AC/CE$. Hence, \[\frac{AB}{BE} = \frac{AC}{CE} = \frac{AB + AC}{BE + CE} = \frac{AB + AC}{BC} = \frac{7 + 8}{9} = \frac{15}{9} = \boxed{\frac{5}{3}}.\]

(Where did E come from?????)

--vaporwave

See Also

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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