# Difference between revisions of "2002 AMC 12A Problems/Problem 23"

## Problem

In triangle $ABC$ , side $AC$ and the perpendicular bisector of $BC$ meet in point $D$, and $BD$ bisects $. If $AD=9$ and $DC=7$, what is the area of triangle ABD?

$\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5$

## Solution

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Looking at the triangle BCD, we see that its perpendicular bisector reaches the vertex, therefore hinting it is isoceles. Let angle C be x. B=2x from given and the previous deducted. <ABD=x, <ADB=2x (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 ABD and ACB are similar.

$\frac {16}{AB}=\frac {AB}{9}$ $AB=12$

Then by using Heron's Formula on ABD (12,7,9 as sides), we have $\sqrt{14(2)(7)(5)}$ $14\sqrt5=D$