Example 12. In convex quadrilateral $A B C D$, $A B=A C$ $=B D, \angle C A B=36^{\circ}, \angle D B A=48^{\circ}$.

Find: $\angle B C D, \angle A D C$.
Analyze: Draw $B H \perp A D$  
at $H$ intersecting $A C$ at $S$, connect $D S$, extend $B C$ to $E$. It is easy to know that
$$
\begin{array}{l}
\angle A B H=\angle D B H= \\
\angle D B C=24^{\circ}, \angle A S H= \\
\angle C A B+\angle A B H=36^{\circ}+ \\
24^{\circ}=60^{\circ} \text{. Thus } \angle D S H=\angle A S H=60^{\circ}, \angle D S C \\
=180^{\circ}-\angle A S D=180^{\circ}-60^{\circ} \times 2=90^{\circ} . \text{ Therefore, }
\end{array}
$$

$S D$ is an angle bisector of $\triangle S B C$. Hence, $B D$ and $C D$ must be parallel to the exterior angle $\angle S C E$ of $\triangle S E C \Rightarrow$ $\angle \mathrm{SCD}=\frac{1}{2} \angle O C E=\frac{1}{2}\left(180^{\circ}-\angle A C B\right)=\frac{1}{2}$ - $\left(180^{\circ}-72^{\circ}\right)=54^{\circ}$, so $\angle B C D=72^{\circ}+54^{\circ}=$ 126 $^{\circ} \angle A D C=360^{\circ}-\angle D A B-\angle A B C-$ $\angle B C D=360^{\circ}-66^{\circ}-72^{\circ}-126^{\circ}=96^{\circ}$.