1983 AHSME Problems/Problem 28
Problem 28
Triangle in the figure has area . Points and , all distinct from and , are on sides and respectively, and . If triangle and quadrilateral have equal areas, then that area is
Solution
Let be the intersection point of and . Since , we have , i.e. . It therefore follows that , so . Now, taking as the base of both and , and using the fact that triangles with the same base and same perpendicular height have the same area, we deduce that the perpendicular distance from to is the same as the perpendicular distance from to . This in turn implies that , and so as , , and are collinear, . Thus , so . Since and have the same perpendicular height (taking as the base), , and hence the answer is .
See Also
1983 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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