- 在 $\Delta ABC$ 及 $\Delta AED$，
$\begin{array}{rcll}
AB & = & AE & \text{(已知)}\
\angle ABC & = & \angle AED & \text{(已知)} \\
\end{array}$另外，
$\begin{array}{rcll}
\angle DAB & = & 180^\circ -\angle ABC & \text{(同旁內角，$AD$//$BC$)} \\
\angle DAB & = & 180^\circ -\angle AED & \text{(已證)} \\
\angle DAB & = & \angle CAE & \text{(同旁內角，$AC$//$ED$)} \\
\angle DAC +\angle CAB & = & \angle DAC +\angle DAE \\
\angle CAB & = & \angle DAE
\end{array}$$\therefore \Delta ABC \cong \Delta AED$ (A.S.A.)。
-
$\begin{array}{rcll}
\angle ACB & = & 180^\circ -\angle BAC -\angle ABC & \text{(三角形的內角和)} \\
\angle ACB & = & 180^\circ -\angle EAD -\angle ABC & \text{($\cong\Delta$ 的對應角)} \\
\angle ACB & = & 180^\circ -87^\circ -39^\circ \\
\angle ACB & = & 54^\circ
\end{array}$另外，
$\begin{array}{rcll}
\angle CAD & = & \angle ACB & \text{(錯角，$AD$//$BC$)} \\
\angle CAD & = & 54^\circ
\end{array}$在 $\Delta ACD$ 中，
$\begin{array}{rcll}
AC & = & AD & \text{($\cong \Delta$ 的對應邊)} \\
\therefore \angle ACD & = & \angle ADC & \text{(等腰三角形的底角)} \\
\angle ACD & = & \dfrac{1}{2}(180^\circ -\angle CAD) & \text{(三角形的內角和)} \\
\angle ACD & = & \dfrac{1}{2}(180^\circ -54^\circ) \\
\angle ACD & = & 63^\circ
\end{array}$
2022-I-08
答案：(b) $63^\circ$
