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 the angle bisector theorem, we have $$\frac{AC}{CG} = \frac{AB}{BG}$$
$$\implies \frac{10}{CG} = \frac{50}{BG}$$
$$\implies BG = 5 \cdot CG$$
Now, note that the ratio of $[ABG]$ and $[AGC]$ is equivalent to the ratio of $BG$ and $CG$. Since the total area is $120$, we get $$[ABG] = 120 \cdot \frac{5}{6} = 100$$ Also, note that since $\triangle ADF$ is similar to $\triangle ABG$ by a factor of $2$, we have $$[AFD] = [ABG] \cdot \frac{1}{4} = 100 \cdot \frac{1}{4} = 25$$
So, $$[FDBG] = [ABG] - [ADF] = 100 - 25 = 75 \longrightarrow \boxed{\textbf{(D)}}$$
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