Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12$.  Diagonals $AC$ and $BD$ intersect at $E$, $AC = 14$, and $\triangle AED$ and $\triangle BEC$ have equal areas.  What is $AE$?
$\textbf{(A)}\ \frac {9}{2}\qquad \textbf{(B)}\ \frac {50}{11}\qquad \textbf{(C)}\ \frac {21}{4}\qquad \textbf{(D)}\ \frac {17}{3}\qquad \textbf{(E)}\ 6$
Let $[ABC]$ denote the area of triangle $ABC$. $[AED] = [BEC]$, so $[ABD] = [AED] + [AEB] = [BEC] + [AEB] = [ABC]$. Since triangles $ABD$ and $ABC$ share a base, they also have the same height and thus $\overline{AB}||\overline{CD}$ and $\triangle{AEB}\sim\triangle{CED}$ with a ratio of $3: 4$. $AE = \frac {3}{7}\times AC$, so $AE = \frac {3}{7}\times 14 = 6\ \boxed{\textbf{(E)}}$.