解：$\because AB = AC，$$\therefore \angle ABC=\angle ACB。$
$\because \angle A = 92^{\circ}，$$\therefore \angle ABC=\angle ACB=\frac{180^{\circ}-92^{\circ}}{2}=44^{\circ}。$
$\because BD = BC，$$\therefore \angle D=\angle BCD。$
$\because \angle ABC=\angle D+\angle BCD = 44^{\circ}，$$\therefore \angle D=\angle BCD = 22^{\circ}。$
$\therefore \angle ACD=\angle BCD+\angle ACB=22^{\circ}+44^{\circ}=66^{\circ}$
