1997 AHSME Problems/Problem 26
Contents
Problem
Triangle and point in the same plane are given. Point is equidistant from and , angle is twice angle , and intersects at point . If and , then
Solution
The product of two lengths with a common point brings to mind the Power of a Point Theorem.
Since , we can make a circle with radius that is centered on , and both and will be on that circle. Since , we can see that point will also lie on the circle, since the measure of arc is twice the masure of inscribed angle , which is true for all inscribed angles.
Since is a line, we have , which gives , or .
We now extend radius to diameter . Since is a line, we have , which gives , or .
Finally, we apply the power of a point theorem to point . This states that . Since and , the desired product is , which is .
Solution 2
Construct the angle bisector of and let it intersect at From the angle bisector theorem, we have and for some Then, note that so is cyclic. Then, or Thus, or
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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