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

By the Angle Bisector Theorem, $\frac{13}{BE} = \frac{15}{14 - BE} \Longrightarrow BE = 6.5$ . Since $BD = 7$ , then $DE = 0.5$ .

By Heron's Formula, $[ABC] = \sqrt{21(6)(7)(8)} = 84$ , so the height of $\triangle ABC$ from $A$ is $h = \frac{2 \cdot 84}{14} = 12$ . Notice that the heights of $\triangle ABC$ and $\triangle ADE$ are the same, so $[ADE] = \frac{1}{2}(12)(0.5) = 3\ \mathrm{(C)}$ .

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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