2002 AMC 12A Problems/Problem 23

Problem

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

$A) 14$ $B) 21$ $C)28$ $D)14\sqrt5$ $E)28\sqrt5$

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. 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 (outer 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 $\sqrt{14(2)(7)(5)}$ $14\sqrt5=E$

2002 AMC 12A (Problems • Answer Key • Resources)
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2002 AMC 12A Problems/Problem 23

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