Triangle $ABC$ is an isosceles triangle with $\overline{AB}=\overline{BC}$. Point $D$ is the midpoint of both $\overline{BC}$ and $\overline{AE}$, and $\overline{CE}$ is 11 units long. Triangle $ABD$ is congruent to triangle $ECD$. What is the length of $\overline{BD}$? 

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.5\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 5.5\qquad\textbf{(E)}\ 6$
Since triangle $ABD$ is congruent to triangle $ECD$ and $\overline{CE} =11$, $\overline{AB}=11$. Since $\overline{AB}=\overline{BC}$,  $\overline{BC}=11$. Because point $D$ is the midpoint of $\overline{BC}$, $\overline{BD}=\frac{\overline{BC}}{2}=\frac{11}{2}=\boxed{\textbf{(D)}\ 5.5}$.