6. As shown in the figure, in $\triangle A B C$, $A B=30, A C=20, S_{\triangle A B C}=210$, $D, E$ are the midpoints of sides $A B, A C$ respectively, and the angle bisector of $\angle B A C$ intersects $D E, B C$ at points $F, G$ respectively, then the area of quadrilateral $B G F D$ is . $\qquad$
By the Angle Bisector Theorem, $\frac{C G}{G B}=\frac{A C}{A B}=\frac{2}{3}$, then $S_{\triangle A B G}=\frac{3}{5} \cdot 210=126$, so $S_{B G F D}=126 \cdot \frac{3}{4}=\frac{189}{2}$.