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 $\angle ABC$ . 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

$[asy] draw((0,0)--(10,0),black); draw((10,0)--(10.5,5)--(0,0),black); draw((10,0)--(5,2.4),black); draw((5,2.4)--(5,0),black); draw((4.6,0)--(4.6,0.4)--(5.4,0.4)--(5.4,0),red); label("A",(10.5,5),N); label("D",(5,2.4),N); label("C",(0,0),W); label("B",(10,0),E); label("9",(7.5,3.7),N); label("7",(2.5,1.6),N); [/asy]$

Looking at the triangle $BCD$ , we see that its perpendicular bisector reaches the vertex, therefore hinting it is isoceles. Let $x = \angle C$ , so that $B=2x$ from given and the previous deducted. Then $\angle ABD=x, \angle ADB=2x$ 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 $\triangle ABD$ and $\triangle ACB$ are similar, so $\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12$ .

Then by using Heron's Formula on $ABD$ (with sides $12,7,9$ ), we have $[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$ .

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2002 AMC 12A Problems/Problem 23

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