2016-II-17

答案：A
由於 $ABCD$ 為一平行四邊形，$AD\text{//}BC$。則可得

$\begin{array}{rcll}\mathrm{\angle }ADC+\mathrm{\angle }BCD& =& {180}^{\circ }& \text{(同旁內角，}AD\text{//}BC\text{)}\\ {114}^{\circ }+\mathrm{\angle }BCD& =& {180}^{\circ }& \\ \mathrm{\angle }BCD& =& {66}^{\circ }& \end{array}$

由於 $BE=CE$，則可得

$\begin{array}{rcll}\mathrm{\angle }EBC& =& \mathrm{\angle }ECB& \text{(等腰 }\mathrm{\Delta }\text{ 的底角)}\\ \mathrm{\angle }EBC& =& {66}^{\circ }& \end{array}$

在 $\mathrm{\Delta }EBC$，

$\begin{array}{rcll}\mathrm{\angle }BEC& =& {180}^{\circ }–\mathrm{\angle }EBC–\mathrm{\angle }ECB& \text{(}\mathrm{\Delta }\text{ 的內角和)}\\ \mathrm{\angle }BEC& =& {180}^{\circ }–{66}^{\circ }–{66}^{\circ }& \\ \mathrm{\angle }BEC& =& {48}^{\circ }\end{array}$

由於 $ABCD$ 為一平行四邊形，$AB\text{//}CD$。則可得

$\begin{array}{rcll}\mathrm{\angle }ABE& =& \mathrm{\angle }BEC& \text{(錯角，}AB\text{//}CD\text{)}\\ \mathrm{\angle }ABE& =& {48}^{\circ }\end{array}$