Ans: $x=70^\circ$, $y=20^\circ$, $z=20^\circ$
Since $DEF$ is a straight line, then
Since $DEF$ is a straight line, then
$\begin{array}{rcl}
110^\circ + x & = & 180^\circ \\
x & = & 70^\circ
\end{array}$
In $\Delta BED$,
$\begin{array}{rcl}
70^\circ + 90^\circ + y & = & 180^\circ \\
y & = & 20^\circ
\end{array}$
Since $AC//DF$, we have
$\begin{array}{rcl}
\angle CBF & = & y \\
\angle CBF & = & 20^\circ
\end{array}$
In $\Delta BCF$, since $BC = CF$, we have
$\begin{array}{rcl}
\angle CFB & = & \angle CBF \\
z & = & 20^\circ
\end{array}$
