1953 AHSME Problems/Problem 28
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Problem 28
In , sides and are opposite and respectively. bisects and meets at . Then if and the correct proportion is:
Solution
By the Angle Bisector Theorem, . Because of this, must equal , where is some real number. Therefore, . Factoring out, we get . However, , so the answer is .
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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