2016-II-17

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

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

由於 $BE=CE$，則可得

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

在 $\Delta EBC$，

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

由於 $ABCD$ 為一平行四邊形，$AB\text{//}CD$。則可得

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