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 = \angle ADC</math> because they subtend the same arc. Furthermore, <math>\angle | + | Let <math>E = \overline{BC}\cap \overline{AD}</math>. Observe that <math>\angle ABC = \angle ADC</math> because they subtend the same arc. Furthermore, <math>\angle BAE = \angle EAC</math>, so <math>\triangle ABE</math> is similar to <math>\triangle ADC</math> by AAA similarity. Then <math>\dfrac{AD}{AB} = \dfrac{CD}{BE}</math>. By angle bisector theorem, <math>\dfrac{7}{BE} = \dfrac{8}{CE}</math> so <math>\dfrac{7}{BE} = \dfrac{8}{9-BE}</math> which gives <math>BE = \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>. |
== See Also == | == See Also == |
Revision as of 12:18, 22 November 2016
In triangle we have , , . Point is on the circumscribed circle of the triangle so that bisects angle . What is the value of ?
Solution
Set 's length as . 's length must also be since and intercept arcs of equal length(because ). Using Ptolemy's Theorem, . The ratio is
Solution 2
Let . Observe that because they subtend the same arc. Furthermore, , so is similar to by AAA similarity. Then . By angle bisector theorem, so which gives . Plugging this into the similarity proportion gives: .
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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