2018-II-22

$\begin{array}{rcll}BD& =& DE& \text{(given)}\\ \mathrm{\angle }DBE& =& \mathrm{\angle }DEB& \text{(base }\mathrm{\angle }\text{s, isos. }\mathrm{\Delta }\text{)}\end{array}$

Then, we have

$\begin{array}{rcll}\mathrm{\angle }DBE+\mathrm{\angle }DEB& =& \mathrm{\angle }ADB& \text{(ext. }\mathrm{\angle }\text{ of }\mathrm{\Delta }\text{)}\\ 2\mathrm{\angle }DEB& =& \mathrm{\angle }ADB\end{array}$

Consider quadrilateral $ABCD$.

$\begin{array}{rcll}\mathrm{\angle }DAC& =& \mathrm{\angle }DBE& \text{(}\mathrm{\angle }\text{s in the same segment)}\\ \mathrm{\angle }DAC& =& \mathrm{\angle }DEB& \text{(proved)}\end{array}$

Consider $\mathrm{\Delta }ABD$.

$\begin{array}{rcll}\mathrm{\angle }ABD+\mathrm{\angle }BAD+\mathrm{\angle }ADB& =& {180}^{\circ }& \text{(}\mathrm{\angle }\text{ sum of }\mathrm{\Delta }\text{)}\\ \mathrm{\angle }ABD+\mathrm{\angle }BAC+\mathrm{\angle }CAD+\mathrm{\angle }ADB& =& {180}^{\circ }\\ {30}^{\circ }+{66}^{\circ }+\mathrm{\angle }DEB+2\mathrm{\angle }DEB& =& {180}^{\circ }& \text{(proved)}\\ 3\mathrm{\angle }DEB& =& {84}^{\circ }\\ \mathrm{\angle }DEB& =& {28}^{\circ }\\ \mathrm{\angle }CED& =& {28}^{\circ }\end{array}$