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

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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 (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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