Ans: $x=33^\circ$, $y=76^\circ$, $z=28^\circ$
Since $AB//CD$, then
Since $AB//CD$, then
$\begin{array}{rcl}
\angle ADC & = & \angle DAB \\
x & = & 33^\circ
\end{array}$
Consider $\Delta CDE$,
$\begin{array}{rcl}
\angle CEA & = & \angle ECD + \angle EDC \\
y & = & 43^\circ + x \\
& = & 43^\circ + 33^\circ \\
&= & 76^\circ
\end{array}$
Consider $\Delta ACE$, since $AE = AC$, we have
$\begin{array}{rcl}
\angle ACE & = & \angle AEC \\
& = & y \\
& = & 76^\circ
\end{array}$
Therefore, we have
$\begin{array}{rcl}
\angle CAE & = & 180^\circ – \angle ACE – \angle AEC \\
z & = & 180^\circ – 76^\circ – 76^\circ \\
& = & 28^\circ
\end{array}$
