2002 AMC 12A Problems/Problem 23
Problem
In triangle , side and the perpendicular bisector of meet in point , and bisects . If and , what is the area of triangle ABD?
Solution
Solution 1 Looking at the triangle , we see that its perpendicular bisector reaches the vertex, therefore implying it is isosceles. Let , so that from given and the previous deducted. Then because any exterior angle of a triangle has a measure that is the sum of the two interior angles that are not adjacent to the exterior angle. That means and are similar, so .
Then by using Heron's Formula on (with sides ), we have .
Solution 2
Let M be the point of the perpendicular bisector on BC. By the perpendicular bisector theorem, and . Also, by the angle bisector theorem, . Thus, let and . In addition, .
Thus, . Additionally, using the Law of Cosines and the fact that ,
Substituting and simplifying, we get
Thus, . We now know all sides of . Using Heron's Formula on ,
Solution 3
Note that because the perpendicular bisector and angle bisector meet at side and as triangle is isosceles, so . By the angle bisector theorem, we can express and as and respectively. We try to find through Stewart's Theorem. So
We plug this to find that the sides of are . By Heron's formula, the area is . ~skyscraper
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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All AMC 12 Problems and Solutions |
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