smc

Looking at the triangle $BCD$, we see that its perpendicular bisector reaches the vertex, therefore implying it is isosceles. Let $x=\angle C$, so that $B=2x$ from given and the previous deducted. Then $\angle ABD=x,\angle ADB=2x$ because any exterior angle of a triangle has a measure that is the sum of the two interior angles that are not adjacent to the exterior angle. That means $△ABD$ and $△ACB$ are similar, so $\frac{16}{AB}=\frac{AB}{9}⟹AB=12$. Then by using Heron's Formula on $ABD$ (with sides $12,7,9$), we have $[△ABD]=\sqrt{14(2)(7)(5)}=14\sqrt{5}⟹\boxed{14\sqrt{5}}$.