2020-I-08

$\begin{array}{rcll}
AB & = & BE & \text{(given)} \\
\angle AEB & = & \angle BAE & \text{(base $\angle$s, isos. $\Delta$)} \\
\angle AEB & = & 30^\circ
\end{array}$

Since $BD//CE$, we have

$\begin{array}{rcll}
\angle AEC & = & \angle ADB & \text{(corr. $\angle$s, $BD//CE$)} \\
\angle AEC & = & 42^\circ
\end{array}$

Therefore, we have

$\begin{array}{rcl}
\angle BEC & = & \angle AEC – \angle AEB \\
\angle BEC & = & 42^\circ – 30^\circ \\
\angle BEC & = & 12^\circ
\end{array}$

$\begin{array}{rcll}
\angle DCE & = & \angle BDC & \text{(alt. $\angle$s, $BD//CE$)} \\
\angle DCE & = & \theta
\end{array}$

In $\Delta CEF$,

$\begin{array}{rcll}
\angle CFE & = & 180^\circ – \angle CEF – \angle ECF & \text{($\angle$ sum of $\Delta$)} \\
\angle CFE & = & 180^\circ – 12^\circ – \theta \\
\angle CFE & = & 168^\circ – \theta
\end{array}$