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 use either Heron's on the larger triangle to calculate the height of ADE or we can decompose ABC into two right triangles of 5-12-13 and 9-12-15 to notice that height is 12. Then angle bisector theorem leads directly to the base of ADE, which leads to an area of 3.

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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2000 AMC 12 Problems/Problem 19

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