6. In convex quadrilateral $A B C D$ we have $A B=15, B C=16, C D=12, D A=25$, and $B D=20$. Let $M$ and $\gamma$ denote the circumcenter and circumcircle of $\triangle A B D$. Line $C B$ meets $\gamma$ again at $F$, line $A F$ meets $M C$ at $G$, and line $G D$ meets $\gamma$ again at $E$. Determine the area of pentagon $A B C D E$.
Answer: 396 $\square$
Note that $\angle A D B=\angle D C B=90^{\circ}$ and $B C \| A D$. Now by Pascal theorem on $D D E B F A$ implies that $B, M, E$ are collinear. So $[A D E]=[A B D]=150$ and $[B C D]=96$, so the total area is 396 .