# 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 1

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 $\frac{12*\frac{1}{2}}{2}=3$.

## Solution 2

The area of $ADE$ is $\frac{DE\cdot h}{2}=\frac{DE}{BC} \cdot \frac{BC\cdot h}{2}=\frac{DE}{BC}[ABC]$ where $h$ is the height of triangle $ABC$. Using Angle Bisector Theorem, we find $\frac{13}{BE}=\frac{15}{14-BE}$, which we solve to get $BE=\frac{13}{2}$. $D$ is the midpoint of $BC$ so $BD=7$. Thus we get the base of triangle $ADE, DE$, to be $\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\sqrt{(21)(8)(7)(6)}=84$$ $$\frac{DE}{BC}[ABC]=\frac{\frac{1}{2}}{14}\cdot 84=3$$ Therefore, the answer is $\mathrm{C}$.

## Solution 3

pair A,B,C,D,E;
B=(0,0);
C=(14;0);
A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180));
draw(A--B--C--cycle);
D=(7,0);
E=(6.5,0);
draw(A--E);
draw(A--D);
label("$A$",A,N);
label("$B$",B,S);
label("$C$",C,S);
label("$E$",E,S);
label("$D$",D,S);
label("$13$",A--B,NW);
label("$15$",A--C,NE);
label("$14$",B--C,S);
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Let's find the area of $\Delta ABC$ by Heron,

$s=\frac{a+b+c}{2}\\\\s=\frac{14+15+13}{2}\to\boxed{s=21}$

Then,

$A^2=s(s-a)(s-b)(s-c)\\\\A^2=21(21-14)(21-15)(21-13)\\\\\boxed{A=84}$

Knowing that D is the midpoint of BC, thus the area of $\Delta ADC=\Delta ABD$, which is half the area of $\Delta ABC$, so $\Delta ADC=\Delta ABD=42$.

By Angle Bisector Theorem we know that:

$\frac{13}{BE}=\frac{15}{CE}$

$\boxed{CE=14-BE}$

$\frac{13}{BE}=\frac{15}{14-BE}$

$\boxed{BE=6.5}\Rightarrow CE=7.5$

Also, we know that:

$\frac{A_{\Delta ADE}}{A_{\Delta ABC}}=\frac{ED}{BC}$

And, we can easily see that $DE=0.5$, so,

$\frac{A_{\Delta ADE}}{84}=\frac{0.5}{14}$

$\boxed{A_{\Delta ADE}=3}$
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