2020-I-08

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

Since $BD//CE$, we have

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

Therefore, we have

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

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

In $\mathrm{\Delta }CEF$,

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