In triangle $ABC$, side $AC$ and the [perpendicular bisector](https://artofproblemsolving.com/wiki/index.php/Perpendicular_bisector) of $BC$ meet in point $D$, and $BD$ bisects $\angle ABC$. If $AD=9$ and $DC=7$, what is the area of triangle $ABD$? 
$\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5$
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 $\triangle ABD$ and $\triangle ACB$ are [similar](https://artofproblemsolving.com/wiki/index.php/Similarity_(geometry)), so $\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12$.
Then by using [Heron's Formula](https://artofproblemsolving.com/wiki/index.php/Heron%27s_Formula) on $ABD$ (with sides $12,7,9$), we have $[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$.