2020-I-08

$\begin{array}{rcll}
AB & = & BE & \text{(已知)} \\
\angle AEB & = & \angle BAE & \text{(等腰 $\Delta$ 的底角)} \\
\angle AEB & = & 30^\circ
\end{array}$

由於 $BD//CE$，可得

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

所以，可得

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

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

在 $\Delta CEF$ 中，

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