Triangle $ABC$ with $AB=60$ and $AC=20$ has area $160$. 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 $FDBG$?

1. $BG:GC=AB:AC=3:1$, so $[ABG]=\frac34[ABC]=\frac34(160)=120$.
2. Since $D,E$ are midpoints, $[FDBG]=\frac34[ABG]=\frac34(120)=\boxed{90}$.