Example 6 As shown in Figure 9, in the convex quadrilateral $ABCD$, $AB = AC = BD$. Among its four interior angles, two are acute angles, with degree measures of $72^{\circ}$ and $66^{\circ}$. Find the degree measures of the other two interior angles.
Obviously, $\triangle A B C$ and $\triangle B A D$ are both isosceles triangles. Since the base angles of an isosceles triangle are acute, we know that $\angle B A D$ and $\angle A B C$ are both acute angles. Let's assume $\angle B A D = 72^{\circ}$ and $\angle A B C = 66^{\circ}$. At this point,
$$
\begin{array}{l}
\angle B D A = 72^{\circ}, \angle A C B = 66^{\circ}, \\
\angle B A C = 180^{\circ} - 2 \times 66^{\circ} = 48^{\circ}, \\
\angle A B D = 180^{\circ} - 2 \times 72^{\circ} = 36^{\circ}.
\end{array}
$$

As shown in Figure 9, draw $A H \perp B C$ at point $H$, and let it intersect $B D$ at point $S$. Connect $S C$ and extend $A D$ to point $K$. It is easy to see that
$$
\begin{array}{l}
\angle C A D = \angle B A D - \angle B A C = 24^{\circ}, \\
\angle S A C = \angle H A C = \frac{1}{2} \angle B A C = 24^{\circ}.
\end{array}
$$

Thus, $\angle S A C = \angle C A D$, which means $A C$ bisects $\angle D A S$.
Also, $\angle B S H = 90^{\circ} - \angle S B H$
$$
= 90^{\circ} - (\angle A B C - \angle A B D) = 60^{\circ},
$$

It is easy to see that $\angle C S H = \angle B S H = 60^{\circ}$. Therefore,
$$
\angle D S C = 180^{\circ} - (\angle B S H + \angle C S H) = 60^{\circ}.
$$

Thus, $\angle D S C = \angle C S H$. This means $S C$ bisects the exterior angle of $\angle A S D$.

Since $A C$ and $S C$ bisect $\angle D A S$ and the exterior angle of $\angle A S D$ respectively, $C$ must be the excenter of $\triangle A S D$.

According to the properties of the excenter, $D C$ bisects the exterior angle of $\angle A D S$.
Since $\angle B D A = 72^{\circ}$ and $\angle B D K = 108^{\circ}$, then
$$
\begin{array}{l}
\angle B D C = \angle S D C = \frac{1}{2} \angle B D K = 54^{\circ}, \\
\angle A D C = \angle B D A + \angle B D C = 126^{\circ}.
\end{array}
$$

Therefore, $\angle B C D = 360^{\circ} - 72^{\circ} - 66^{\circ} - 126^{\circ} = 96^{\circ}$.