Ans: C
In $\Delta CDE$, since $CE=DE$, we have
In $\Delta CDE$, since $CE=DE$, we have
$\begin{array}{rcl}
\angle EDC & = & \angle ECD \\
& = & 20^\circ
\end{array}$
Since $ABCD$ is a parallelogram, then $AB//CD$. Hence, we have
$\begin{array}{rcl}
\angle DBA & = & \angle BDC \\
& = & 20^\circ
\end{array}$
In $\Delta ABE$,
$\begin{array}{rcl}
\angle EBA + \angle BAE & = & \angle AED \\
20^\circ + \angle BAE & = & 130^\circ \\
\angle BAE & = & 110^\circ
\end{array}$
