由於 $AB=AD$,可得
$\begin{array}{ll}
\angle ABD = \angle ADB & \text{(等腰 $\Delta$ 的底角)} \\
\angle ABD = 58^\circ
\end{array}$
由於 $ABCD$ 為圓內接四邊形,可得
$\begin{array}{ll}
\angle ABC + \angle ADC = 180^\circ & \text{(圓內接四邊形的對角)} \\
\end{array}$
$\begin{array}{rcl}
\angle DBC & = & 180^\circ – \angle ADB – \angle ABC – \angle CBD \\
\angle DBC & = & 180^\circ – 58^\circ – 58^\circ – 25^\circ \\
\angle DBC & = & 39^\circ
\end{array}$
留意
$\begin{array}{ll}
\angle ACB = \angle ADB & \text{(同弓形內的圓周角)} \\
\angle ACB = 58^\circ
\end{array}$
在 $\Delta BCE$ 中,由於 $BC=CE$,可得
$\begin{array}{ll}
\angle CBE = \angle CEB & \text{(等腰 $\Delta$ 的底角)}
\end{array}$
所以,可得
$\begin{array}{ll}
\angle CBE = (180^\circ – \angle BCE) \div 2 & \text{($\Delta$ 的內角和)} \\
\angle CBE = (180^\circ – 58^\circ) \div 2 \\
\angle CBE = 61^\circ
\end{array}$
所以,$\angle ABE$
$\begin{array}{cl}
= & \angle ABC – \angle CBE \\
= & 58^\circ + 25^\circ – 61^\circ \\
= & 22^\circ
\end{array}$