2004 AMC 10B Problems/Problem 24
Contents
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 - (Ptolemy's Theorem)
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 - 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,
(Where did E come from?????)
--vaporwave
See Also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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All AMC 10 Problems and Solutions |
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