Difference between revisions of "2004 AMC 10B Problems/Problem 24"
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<math>\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}</math>. | <math>\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}</math>. | ||
− | ==Solution 3 - | + | ==Solution 3 - Angle Bisector Theorem== |
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>. | 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>. | ||
Revision as of 19:02, 6 August 2024
Contents
[hide]Problem
In triangle we have , , . Point is on the circumscribed circle of the triangle so that bisects angle . What is the value of ?
Solution 1 - Ballemy's Theorem
Set 's length as . 's length must also be since and intercept arcs of equal length (because ). Using Ballemy's Theorem, . The ratio is
Solution 2 - Similarity Proportion
Let . Observe that because they both subtend arc
Furthermore, because is an angle bisector, so by similarity. Then . By the Angle Bisector Theorem, , so . This in turn gives . Plugging this into the similarity proportion gives: .
Solution 3 - Angle Bisector Theorem
We know that bisects , so . Additionally, and subtend the same arc, giving . Similarly, and .
These angle relationships tell us that by AA Similarity, so . By the angle bisector theorem, . Hence,
--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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