smc

Let $BC=a$, $BG=x$, $GC=y$, and the length of the perpendicular from $BC$ through $A$ be $h$. By angle bisector theorem, we have that $$\frac{50}{x}=\frac{10}{y},$$ where $y=-x+a$. Therefore substituting we have that $BG=\frac{5a}{6}$. By similar triangles, we have that $DF=\frac{5a}{12}$, and the height of this trapezoid is $\frac{h}{2}$. Then, we have that $\frac{ah}{2}=120$. We wish to compute $\frac{5a}{8}\cdot \frac{h}{2}$, and we have that it is $\boxed{75}$ by substituting.