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
![[asy] unitsize(0.25 cm); pair A, B, C, D, M; A = (0,0); B = (88/9, 28*sqrt(5)/9); C = (16,0); D = 9/16*C; M = (B + C)/2; draw(A--B--C--cycle); draw(B--D--M); label("$A$", A, SW); label("$B$", B, N); label("$C$", C, SE); label("$D$", D, S); [/asy]](http://latex.artofproblemsolving.com/4/d/e/4deb1610580c4eae6c5d06124365b18655edbd9a.png)
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 ![$[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$](http://latex.artofproblemsolving.com/b/8/3/b83147bdfd7a18000dfc73b2e8f65244b6a7e326.png)
.
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 , 
See Also
| 2002 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 22 |
Followed by Problem 24 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
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