2006-I-05

$\begin{array}{rcl}\mathrm{\angle }AEB& =& \mathrm{\angle }CBE\\ & =& {70}^{\circ }\end{array}$

因為 $AE=AB$，則可得

$\begin{array}{rcl}\mathrm{\angle }ABE& =& \mathrm{\angle }AEB\\ & =& {70}^{\circ }\end{array}$

在 $\mathrm{\Delta }ABE$ 中，

$\begin{array}{rcl}\mathrm{\angle }BAC& =& {180}^{\circ }–\mathrm{\angle }ABE–\mathrm{\angle }AEB\\ & =& {180}^{\circ }–{70}^{\circ }–{70}^{\circ }\\ & =& {40}^{\circ }\end{array}$

因為 $ABCD$ 為平行四邊形，則可得

$\begin{array}{rcl}\mathrm{\angle }BCD& =& \mathrm{\angle }BAD\\ & =& {40}^{\circ }\end{array}$