1998 AHSME Problems/Problem 26
Problem
In quadrilateral , it is given that , angles and are right angles, , and . Then
Contents
[hide]Solution
Solution 1
Let the extensions of and be at . Since , and is a triangle. Also, , so is also a triangle.
Thus , and . By the Pythagorean Theorem on ,
Solution 2
Opposite angles add up to , so is a cyclic quadrilateral. Also, , from which it follows that is a diameter of the circumscribing circle. We can apply the extended version of the Law of Sines on :
By the Law of Cosines on :
So .
See also
1998 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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