Since $ABCD$ is a parallelogram, then $AD\text{//}BC$. Then we have
$\begin{array}{rcll}
\angle ADC + \angle BCD & = & 180^\circ & \text{(int. $\angle$s, $AD$//$BC$)} \\
114^\circ + \angle BCD & = & 180^\circ & \\
\angle BCD & = & 66^\circ & \\
\end{array}$
Since $BE=CE$, then we have
$\begin{array}{rcll}
\angle EBC & = & \angle ECB & \text{(base $\angle$s, isos. $\Delta$)} \\
\angle EBC & = & 66^\circ &
\end{array}$
In $\Delta EBC$,
$\begin{array}{rcll}
\angle BEC & = & 180^\circ – \angle EBC – \angle ECB & \text{($\angle$s sum of $\Delta$)} \\
\angle BEC & = & 180^\circ – 66^\circ – 66^\circ & \\
\angle BEC & = & 48^\circ
\end{array}$
Since $ABCD$ is a parallelogram, then $AB\text{//}CD$. Then we have
$\begin{array}{rcll}
\angle ABE & = & \angle BEC & \text{(alt. $\angle$s, $AB$//$CD$)} \\
\angle ABE & = & 48^\circ
\end{array}$
