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 $\tan \angle BDC$?
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 $\tan \angle BDC = \frac{CD}{BD} = \boxed{2}$.
First, let's draw a diagram to better understand the given information. We have right triangle $ABC$ with $\angle BAC = 90^\circ$.
Let $D$ be the midpoint of $\overline{AC}$.
We are given that $AB = 7$ and $BC = 7 \sqrt{3}$.
[Insert diagram here]
To find $\tan \angle BDC$, we need to find the lengths of $\overline{BD}$ and $\overline{DC}$.
Since $D$ is the midpoint of $\overline{AC}$, we have $AD = DC$.
Using the Pythagorean Theorem, we can find the length of $\overline{BC}$:
$BC^2 = AB^2 + AC^2$
$(7\sqrt{3})^2 = 7^2 + AC^2$
$63 = 49 + AC^2$
$14 = AC^2$
Taking the square root of both sides, we get $AC = \sqrt{14}$.
Since $D$ is the midpoint of $\overline{AC}$, we have $AD = \frac{1}{2} \sqrt{14}$.
Using the Pythagorean Theorem again, we can find the length of $\overline{BD}$:
$BD^2 = AB^2 + AD^2$
$BD^2 = 7^2 + \left(\frac{1}{2} \sqrt{14}\right)^2$
$BD^2 = 49 + \frac{1}{4} \cdot 14$
$BD^2 = 49 + \frac{14}{4}$
$BD^2 = 49 + \frac{7}{2}$
$BD^2 = \frac{98}{2} + \frac{7}{2}$
$BD^2 = \frac{105}{2}$
Taking the square root of both sides, we get $BD = \frac{\sqrt{105}}{2}$.
Now, we can find $\tan \angle BDC$:
$\tan \angle BDC = \frac{BD}{CD}$
Since $AD = DC$, we have $CD = AC - AD = \sqrt{14} - \frac{1}{2} \sqrt{14} = \frac{1}{2} \sqrt{14}$.
Therefore, $\tan \angle BDC = \frac{\frac{\sqrt{105}}{2}}{\frac{1}{2} \sqrt{14}} = \frac{\sqrt{105}}{\sqrt{14}} = \boxed{\sqrt{15}}$.