Join $AC$.
$\begin{array}{ll}
BC = CD & \text{(given)} \\
\overparen{BC} = \overparen{CD} & \text{(eq. chord, eq. arc)} \\
\angle BAC = \angle CAD & \text{(arc and $\angle$ at $\unicode{x2299}^{ce}$ in prop.)}
\end{array}$
Therefore, we have
$\begin{array}{rcl}
\angle CAD & = & \dfrac{1}{2} \times \angle BAD \\
\angle CAD & = & \dfrac{1}{2} \times 58^\circ \\
\angle CAD & = & 29^\circ
\end{array}$
Since $AD$ is a diameter of the circle, then we have
$\begin{array}{ll}
\angle ACD = 90^\circ & \text{($\angle$ in semi-circle)} \\
\end{array}$
In $\Delta ACD$,
$\begin{array}{rcll}
\angle ADC & = & 180^\circ – \angle ACD – \angle CAD & \text{($\angle$ sum of $\Delta$)} \\
\angle ADC & = & 180^\circ – 90^\circ – 29^\circ \\
\angle ADC & = & 61^\circ
\end{array}$
Since $A$, $C$, $D$ and $E$ are points on the circumference, then we have
$\begin{array}{rcll}
\angle AEC & = & \angle ADC & \text{($\angle$s in the same segment)} \\
\angle AEC & = & 61^\circ
\end{array}$
