In rectangle $ABCD$, $AB=6$, $AD=30$, and $G$ is the midpoint of $\overline{AD}$. Segment $AB$ is extended $3$ units beyond $B$ to point $E$, and $F$ is the intersection of $\overline{ED}$ and $\overline{BC}$. What is the area of quadrilateral $BFDG$?

1. $\triangle EFB\sim\triangle EDA$, so with $AB=a, AD=d$, extension $e$: $[EDA]=\frac12 d(a+e)$ and $[EFB]=\left(\frac e{a+e}\right)^2[EDA]$; also $[BGA]=\frac{ad}4$.
2. With $a=6,d=30,e=3$: $[EDA]=\frac12(30)(9)=135$, $[EFB]=\left(\frac39\right)^2(135)=15$, $[BGA]=\frac{180}4=45$.
3. $[BFDG]=135-15-45=\boxed{75}$.