In $\Delta ACD$,
$\begin{array}{rcll}
\because \angle ADC & = & \angle CAD & \text{(given)} \\
\therefore AC & = & CD & \text{(sides opp. eq. $\angle$s)} \\
\end{array}$
Also,
$\begin{array}{rcll}
\angle ACD & = & 180^\circ – \angle CAD – \angle ADC & \text{($\angle$ sum of $\Delta$)} \\
\angle ACD & = & 180^\circ – 40^\circ – 40^\circ \\
\angle ACD & = & 100^\circ
\end{array}$
In $\Delta BCE$,
$\begin{array}{rcll}
BC & = & AC & \text{(equilateral $\Delta$)} \\
AC & = & CD & \text{(proved)} \\
CD & = & CE & \text{(given)} \\
\therefore BC & = & CE \\
\therefore \angle CBE & = & \angle BEC & \text{(base $\angle$s, isos. $\Delta$)} \\
\angle BCE & = & 360^\circ – \angle BCA – \angle ACD – \angle DCE & \text{($\angle$s at a pt.)} \\
\angle BCE & = & 360^\circ – 60^\circ – 100^\circ – 78^\circ & \text{(equilateral $\Delta$)}\\
\angle BCE & = & 122^\circ \\
\angle CBE & = & \dfrac{1}{2} \times (180^\circ – \angle BCE) & \text{($\angle$ sum of $\Delta$)} \\
\angle CBE & = & \dfrac{1}{2} \times (180^\circ – 122^\circ) \\
\angle CBE & = & 29^\circ
\end{array}$
