Each side of $\triangle ABC$ is $12$ units. $D$ is the foot of the perpendicular dropped from $A$ on $BC$, 
and $E$ is the midpoint of $AD$. The length of $BE$, in the same unit, is:
$\textbf{(A)}\ \sqrt{18} \qquad \textbf{(B)}\ \sqrt{28} \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ \sqrt{63} \qquad \textbf{(E)}\ \sqrt{98}$
Note that $\triangle ABC$ is an [equilateral triangle](https://artofproblemsolving.com/wiki/index.php/Equilateral_triangle).  From the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem) (or by using 30-60-90 triangles), $AD = 6\sqrt{3}$.  That means $DE = 3\sqrt{3}$.  Using the Pythagorean Theorem again, $BE = \sqrt{63}$, which is answer choice $\boxed{\textbf{(D)}}$.