In [triangle](https://artofproblemsolving.com/wiki/index.php/Triangle) $ABC$, $AB = 13$, $BC = 14$, $AC = 15$. Let $D$ denote the [midpoint](https://artofproblemsolving.com/wiki/index.php/Midpoint) of $\overline{BC}$ and let $E$ denote the intersection of $\overline{BC}$ with the [bisector](https://artofproblemsolving.com/wiki/index.php/Angle_bisector) of angle $BAC$. Which of the following is closest to the area of the triangle $ADE$?
$\text {(A)}\ 2 \qquad \text {(B)}\ 2.5 \qquad \text {(C)}\ 3 \qquad \text {(D)}\ 3.5 \qquad \text {(E)}\ 4$
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](https://artofproblemsolving.com/wiki/index.php/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 $\frac{12*\frac{1}{2}}{2}=3$.