- In $\Delta ABC$ and $\Delta AED$,
$\begin{array}{rcll}
AB & = & AE & \text{(given)} \\
\angle ABC & = & \angle AED & \text{(given)} \\
\end{array}$On the other hand,
$\begin{array}{rcll}
\angle DAB & = & 180^\circ -\angle ABC & \text{(int. $\angle$s, $AD$//$BC$)} \\
\angle DAB & = & 180^\circ -\angle AED & \text{(proved)} \\
\angle DAB & = & \angle CAE & \text{(int. $\angle$s, $AC$//$ED$)} \\
\angle DAC +\angle CAB & = & \angle DAC +\angle DAE \\
\angle CAB & = & \angle DAE
\end{array}$$\therefore \Delta ABC \cong \Delta AED$ (A.S.A.).
-
$\begin{array}{rcll}
\angle ACB & = & 180^\circ -\angle BAC -\angle ABC & \text{($\angle$ sum of $\Delta$)} \\
\angle ACB & = & 180^\circ -\angle EAD -\angle ABC & \text{(corr. $\angle$s, $\cong\Delta$s)} \\
\angle ACB & = & 180^\circ -87^\circ -39^\circ \\
\angle ACB & = & 54^\circ
\end{array}$Also,
$\begin{array}{rcll}
\angle CAD & = & \angle ACB & \text{(alt. $\angle$, $AD$//$BC$)} \\
\angle CAD & = & 54^\circ
\end{array}$In $\Delta ACD$,
$\begin{array}{rcll}
AC & = & AD & \text{(corr. sides, $\cong \Delta$s)} \\
\therefore \angle ACD & = & \angle ADC & \text{(base $\angle$s, isos. $\Delta$)} \\
\angle ACD & = & \dfrac{1}{2}(180^\circ -\angle CAD) & \text{($\angle$ sum of $\Delta$)} \\
\angle ACD & = & \dfrac{1}{2}(180^\circ -54^\circ) \\
\angle ACD & = & 63^\circ
\end{array}$
2022-I-08
Ans: (b) $63^\circ$
