The ratio of the sides in a $30^\circ$-$60^\circ$-$90^\circ$ triangle is $1:\sqrt{3}:2$, so in triangle $ABC$, $\frac{AB}{BC} = \frac{1}{2}$. Therefore, triangle $ABC$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle.
Since triangle $ABC$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle, $AC = 2AB = 14$. Therefore, $CD = \frac{1}{2} AC = 7$. The opposite side and adjacent side for $\angle BDC$ are $CD$ and $BD = \frac{1}{2} AC$ respectively, so $\tan \angle BDC = \frac{CD}{BD} = \boxed{2}$.
In right triangle $ABC$, we have $\angle BAC = 90^\circ$ and $D$ is the midpoint of $\overline{AC}$. If $AB = 7$ and $BC = 7 \sqrt{3}$, then what is $\tan \angle BDC$?
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