# Difference between revisions of "2018 AMC 10A Problems/Problem 24"

## Problem

Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$. Let $D$ be the midpoint of $\overline{AB}$, and let $E$ be the midpoint of $\overline{AC}$. The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$, respectively. What is the area of quadrilateral $FDBG$?

$\textbf{(A) }60 \qquad \textbf{(B) }65 \qquad \textbf{(C) }70 \qquad \textbf{(D) }75 \qquad \textbf{(E) }80 \qquad$

## Solution

By angle bisector theorem, $BG=\frac{5a}{6}$. By similar triangles, $DF=\frac{5a}{12}$, and the height of this trapezoid is $\frac{h}{2}$, where $h$ is the length of the altitude to $BC$. Then $\frac{ah}{2}=120$ and we wish to compute $\frac{5a}{8}\cdot\frac{h}{2}=\boxed{75}$. (trumpeter)

## (More basic) Solution 2 (bad)

$\overline{DE}$ is midway from $A$ to $\overline{BC}$, and $DE = \frac{BC}{2}$. Therefore, $DE$ is a quarter of the area, which is $30$. Using the angle bisector theorem in the same fashion as the previous problem, we get that one segment is $5$ times the other. We want the larger piece, as described by the problem. Because the heights are identical, one area is $5$ times the other, and $\frac{5}{6} \cdot 90 = \boxed{75}$.