Since $AB=AD$, then we have
$\begin{array}{ll}
\angle ABD = \angle ADB & \text{(base $\angle$s, isos. $\Delta$)} \\
\angle ABD = 58^\circ
\end{array}$
Since $ABCD$ is a cyclic quadrilateral, then we have
$\begin{array}{ll}
\angle ABC + \angle ADC = 180^\circ & \text{(opp. $\angle$s, cyc. quad.)} \\
\end{array}$
$\begin{array}{rcl}
\angle DBC & = & 180^\circ – \angle ADB – \angle ABC – \angle CBD \\
\angle DBC & = & 180^\circ – 58^\circ – 58^\circ – 25^\circ \\
\angle DBC & = & 39^\circ
\end{array}$
Note that
$\begin{array}{ll}
\angle ACB = \angle ADB & \text{($\angle$s in the same segment)} \\
\angle ACB = 58^\circ
\end{array}$
In $\Delta BCE$, since $BC=CE$, we have
$\begin{array}{ll}
\angle CBE = \angle CEB & \text{(base $\angle$s, isos. $\Delta$)}
\end{array}$
Therefore, we have
$\begin{array}{ll}
\angle CBE = (180^\circ – \angle BCE) \div 2 & \text{($\angle$ sum of $\Delta$)} \\
\angle CBE = (180^\circ – 58^\circ) \div 2 \\
\angle CBE = 61^\circ
\end{array}$
Therefore, $\angle ABE$
$\begin{array}{cl}
= & \angle ABC – \angle CBE \\
= & 58^\circ + 25^\circ – 61^\circ \\
= & 22^\circ
\end{array}$
