In triangle $ABC$, side $AC$ and the perpendicular bisector of $BC$ meet in point $D$, and $BD$ bisects $\angle ABC$. If $AD=9$ and $DC=16$, what is the area of triangle $ABD$?

$DB=DC=16$, since $D$ is on the perpendicular bisector of $BC$; hence $\angle DBC=\angle C$.
$\angle ABD=\angle DBC=\angle C$ and $\angle A$ is common, so $\triangle ADB\sim \triangle ABC$.
$AD:AB=AB:AC$, so $AB^2=AD\cdot AC=9\cdot 25$ (as $AC=9+16=25$), and $AB=15$.
$\triangle ADB$ has sides $9,15,16$, semi-perimeter $s=20$, so its area is $\sqrt{20\cdot 11\cdot 5\cdot 4}=\sqrt{4400}=\boxed{20\sqrt{11}}$.