1984 AHSME Problems/Problem 4
Problem
A rectangle intersects a circle as shown: , , and . Then equals:
Solution
Draw and , forming a trapezoid. Since it's cyclic, this trapezoid must be isosceles. Also, drop altitudes from to , to , and to , and let the feet of these altitudes be , , and respectively. is a rectangle since it has right angles. Therefore, , and . By the same logic, is also a rectangle, and . since they're both altitudes to a trapezoid, and since the trapezoid is isosceles. Therefore, by HL congruence, so . Also, is a rectangle from right angles, and . Therefore, .
See Also
1984 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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