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

Line 6: Line 6:
 
== Solution ==
 
== Solution ==
  
Set <math> BD </math>'s length as <math> x </math>. <math> CD </math>'s length must also be <math> x </math>. Using Ptolemy's Theorem, <math> 7x+8x=9(AD) </math>. The ratio is <math> \boxed{\frac{5}{3}}\implies(C) </math>
+
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(C)</math>
 +
 
 +
 
 +
== See Also ==
 +
 
 +
{{AMC10 box|year=2004|ab=B|num-b=23|num-a=25}}

Revision as of 16:57, 1 May 2011

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 $\segment BD$ (Error compiling LaTeX. ! Undefined control sequence.)'s length as $x$. $CD$'s length must also be $x$ since $\angle BAD$ and $\angle DAC$ intercept arcs of equal length. Using Ptolemy's Theorem, $7x+8x=9(AD)$. The ratio is $\boxed{\frac{5}{3}}\implies(C)$


See Also

2004 AMC 10B (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS