2009 AMC 10B Problems/Problem 20
Problem
Triangle has a right angle at , , and . The bisector of meets at . What is ?
Solution 1
By the Pythagorean Theorem, . Then, from the Angle Bisector Theorem, we have:
Solution 2
Let . Notice and . By the double angle identity,
Remarks: You could also use tangent half angle formula
Solution 3
Let .
Make a line so that it is perpendicular to . Since is an angle bisector, is congruent to . Using the Pythagorean Theorem:
We know that by the congruent triangles, so . We know that , , and . We now have right triangle and its 3 sides. Using the Pythagorean Thereom, we get:
So,
~HelloWorld21
Video Solution by OmegaLearn
https://youtu.be/4_x1sgcQCp4?t=4816
~ pi_is_3.14
See Also
2009 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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All AMC 10 Problems and Solutions |
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