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

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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>   
 
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:  <math>\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}</math>.
+
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:   
 +
<math>\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}</math>.
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alternatively:
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<math>\frac{AB}{EB}=\frac{AD}{CD}=\frac{7}{21/5}=7\cdot\frac{5}{21}=\frac{5}{3}</math>
  
 
==Solution 3==
 
==Solution 3==

Revision as of 18:53, 6 August 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

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 Ballemy'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 $\angle ABC \cong \angle ADC$ 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: $\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}$. alternatively: $\frac{AB}{EB}=\frac{AD}{CD}=\frac{7}{21/5}=7\cdot\frac{5}{21}=\frac{5}{3}$

Solution 3

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, $AB/BE = AC/CD$. Hence, \[\frac{AB}{BE} = \frac{AC}{CD} = \frac{7}{21/5} = 7\cdot\frac{5}{21} = \frac{35}{21} = \frac{5}{3}.\]

--vaporwave

P.S We get $AB$ by the numbers given in the problem. We get $BE$ by setting up a systems of equations. Using the Angle Bisector Theorem: $\frac{7}{8}=\frac{EB}{EC}$

We also know that $EB$ and $EC$ add up to 9 (using the numbers given in the problem) $EB+EC=9$

We then solve.

--rosebuddy_vxd

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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