2018-II-22

$\begin{array}{rcll}BD& =& DE& \text{(已知)}\\ \mathrm{\angle }DBE& =& \mathrm{\angle }DEB& \text{(等腰 }\mathrm{\Delta }\text{ 的底角)}\end{array}$

所以，可得

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

考慮四邊形 $ABCD$。

$\begin{array}{rcll}\mathrm{\angle }DAC& =& \mathrm{\angle }DBE& \text{(同弓形內的圓周角)}\\ \mathrm{\angle }DAC& =& \mathrm{\angle }DEB& \text{(已證)}\end{array}$

考慮 $\mathrm{\Delta }ABD$。

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