2000 AMC 12 Problems/Problem 19

Problem

In triangle $ABC$ , $AB = 13$ , $BC = 14$ , $AC = 15$ . Let $D$ denote the midpoint of $\overline{BC}$ and let $E$ denote the intersection of $\overline{BC}$ with the bisector of angle $BAC$ . Which of the following is closest to the area of the triangle $ADE$ ?

$\text {(A)}\ 2 \qquad \text {(B)}\ 2.5 \qquad \text {(C)}\ 3 \qquad \text {(D)}\ 3.5 \qquad \text {(E)}\ 4$

Solution

The answer is exactly $3$ , choice $\mathrm{(C)}$ . We can find the area of triangle $ADE$ by using the simple formula $\frac{bh}{2}$ . Dropping an altitude from $A$ , we see that it has length $12$ ( we can split the large triangle into a $9-12-15$ and a $5-12-13$ triangle). Then we can apply the Angle Bisector theorem on triangle $ABC$ to solve for $BE$ . Solving $\frac{13}{BE}=\frac{15}{14-BE}$ , we get that $BE=\frac{13}{2}$ . $D$ is the midpoint of $BC$ so $BD=7$ . Thus we get the base of triangle ADE, DE, to be $\frac{1}{2}$ units long. Applying the formula $\frac{bh}{2}$ , we get $\frac{12*\frac{1}{2}}{2}=3$ .

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2000 AMC 12 Problems/Problem 19

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