As shown in the figure, in quadrilateral $A B C D$, $\angle B=90^{\circ}, \angle D A B=\angle D=96^{\circ}$, $\angle B C D=78^{\circ}$, and $D A=2 A B$, find $\angle C A B$.
Suppose $A B-1, A D=2$, if the extensions of $D A$ and $C B$ intersect at point $F$ (not shown in the diagram), then $\angle F=6^{\circ}$. It is easy to see that $A F=\frac{1}{\sin 6^{\circ}}, B F=\cot 6^{\circ}$, hence $D F=$ $\frac{1}{\sin 6^{\circ}}+2$. By the Law of Sines, we have $\frac{B C+\cot 6^{\circ}}{D F}=\frac{\sin 96^{\circ}}{\sin 78^{\circ}}$, thus
$$
\begin{aligned}
B C= & =\frac{\sin 96^{\circ}\left(2 \sin 6^{\circ}+1\right)}{\sin 78^{\circ} \sin 6^{\circ}}-\cot 6^{\circ} . \\
& \frac{\sin 96^{\circ}\left(2 \sin 6^{\circ}+1\right)}{\sin 78^{\circ} \sin 6^{\circ}}-\cot 6^{\circ}=\frac{2 \sin 96^{\circ}\left(\sin 6^{\circ}+\sin 30^{\circ}\right)}{\sin 78^{\circ} \sin 6^{\circ}}-\cot 6^{\circ} \\
= & \frac{4 \sin 96^{\circ} \sin 18^{\circ} \cos 12^{\circ}}{\sin 78^{\circ} \sin 6^{\circ}}-\cot 6^{\circ}=\frac{4 \sin 96^{\circ} \sin 18^{\circ}-\cos 6^{\circ}}{\sin 6^{\circ}} \\
= & \left(4 \sin 18^{\circ}-1\right) \cot 6^{\circ}=\left(\frac{2 \sin 36^{\circ}-\cos 18^{\circ}}{\cos 18^{\circ}}\right) \cot 6^{\circ} \\
= & \frac{\cos 54^{\circ}+\cos 54^{\circ}-\cos 18^{\circ}}{\cos 18^{\circ}} \cot 6^{\circ}=\frac{\cos 54^{\circ}\left(1-2 \cos 72^{\circ}\right)}{\cos 18^{\circ}} \cot 6^{\circ} \\
= & \frac{4 \cos 54^{\circ} \sin 6^{\circ} \sin 66^{\circ}}{\cos 18^{\circ}} \cot 6^{\circ}=\frac{4 \cos 6^{\circ} \cos 54^{\circ} \sin 66^{\circ}}{\cos 18^{\circ}} . \\
= & \frac{4 \cos 6^{\circ} \cos 54^{\circ} \cos 66^{\circ}}{\cos 18^{\circ}} \tan 66^{\circ}=\tan 66^{\circ} .
\end{aligned}
$$

Therefore, $\angle C A B=66^{\circ}$.