Join $AC$.
In quadrilateral $ABCD$,
$\begin{array}{rcll}
\angle ADC & = & 180^\circ – \angle ABC & \text{(opp. $\angle$s of cyclic quad.)} \\
\angle ADC & = & 180^\circ – 110^\circ \\
\angle ADC & = & 70^\circ
\end{array}$
Since $AD$ is a diameter, then $\angle ACD = 90^\circ$ ($\angle$ in semi-circle).
In $\Delta ACD$,
$\begin{array}{rcll}
\angle CAD & = & 180^\circ – \angle ACD – \angle ADC & \text{($\angle$ sum of $\Delta$)} \\
\angle CAD & = & 180^\circ – 90^\circ – 70^\circ \\
\angle CAD & = & 20^\circ
\end{array}$
Also,
$\begin{array}{rcll}
BC & = & CD & \text{(given)} \\
\overparen{BC} & = & \overparen{CD} & \text{(eq. chords, eq. arcs)} \\
\angle BAC & = & \angle CAD & \text{(arcs and $\angle$s at $\unicode{x2299}^{ce}$ in prop.)} \\
\angle ABC & = & 20^\circ &
\end{array}$
Hence, we have
$\begin{array}{rcll}
\angle BED & = & \angle BAD & \text{($\angle$s in the same segment)} \\
\angle BED & = & \angle BAC + \angle CAD \\
\angle BED & = & 20^\circ + 20^\circ \\
\angle BED & = & 40^\circ
\end{array}$
