2009 AIME I Problems/Problem 5
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Problem
Triangle has and . Points and are located on and respectively so that , and is the angle bisector of angle . Let be the point of intersection of and , and let be the point on line for which is the midpoint of . If , find .
Solution
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Since is the midpoint of .
Thus, and the opposite angles are congruent.
Therefore, is congruent to because of SAS
is congruent to because of CPCTC
That shows is parallel to (also )
That makes similar to
Thus,
Now lets apply the angle bisector theorem.
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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