# 2004 AMC 12A Problems/Problem 8

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## Problem

In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$, $\angle{EAB}$ and $\angle{ABC}$ are right angles, $AB=4$, $BC=6$, $AE=8$, and $\overline{AC}$ and $\overline{BE}$ intersect at $D$. What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$?

$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$

## Solution

### Solution 1

If we let $[\ldots]$ denote area, $[ABE] - [ABC] = [ADE] + [ABD] - [ABD] - [BDC] = [ADE] - [BDC]$. Using the given, $[ABE] = \frac 12 \cdot 8 \cdot 4$ and $[ABC] = \frac 12 \cdot 6 \cdot 4$, and their difference is $16 - 12 = 4\ \mathrm{(B)}$.

### Solution 2

Since $AE \perp AB$ and $BC \perp AB$, $AE \parallel BC$. By alternate interior angles and AA~, we find that $\triangle ADE \sim \triangle CDB$, with side length ratio $\frac{4}{3}$. Their heights also have the same ratio, and since the two heights add up to $4$, we have that $h_{ADE} = 4 \cdot \frac{4}{7} = \frac{16}{7}$ and $h_{CDB} = 3 \cdot \frac 47 = \frac {12}7$. Subtracting the areas, $\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \frac{12}7 = 4$.