2018 AMC 10A #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 $\mathrm{\angle }BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$, respectively. What is the area of quadrilateral $FDBG$?

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

Solution:

By the angle bisector theorem, we have $$\frac{AC}{CG}=\frac{AB}{BG}$$

$$⟹\frac{10}{CG}=\frac{50}{BG}$$

$$⟹BG=5\cdot CG$$

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 $\mathrm{△}ADF$ is similar to $\mathrm{△}ABG$ by a factor of $2$, we have $$[AFD]=[ABG]\cdot \frac{1}{4}=100\cdot \frac{1}{4}=25$$

So, $$[FDBG]=[ABG]-[ADF]=100-25=75⟶\boxed{{\displaystyle \mathbf{\text{(D)}}}}$$