Example 8 As shown in Figure 11, in quadrilateral $A B C D$,
$$
\begin{array}{l}
\angle C A B=30^{\circ}, \\
\angle A B D=26^{\circ}, \\
\angle A C D=13^{\circ}, \\
\angle D B C=51^{\circ} .
\end{array}
$$

Find the measure of $\angle A D B$.
Solution 1: Let $\angle A D B=x$, then,
$$
\angle A D C=x+43^{\circ} \text {. }
$$

Since $\angle C A B=30^{\circ}, \angle A B D=26^{\circ}$,
$$
\angle D B C=51^{\circ}, \angle A C D=13^{\circ} \text {, }
$$

then $\angle A C B=73^{\circ}, \angle B D C=43^{\circ}$.
Applying the trigonometric form of Ceva's Theorem to $\triangle B C D$ and point $A$ we have
$$
\begin{aligned}
1 & =\frac{\sin \angle D C A}{\sin \angle A C B} \cdot \frac{\sin \angle C B A}{\sin \angle A B D} \cdot \frac{\sin \angle B D A}{\sin \angle A D C} \\
& =\frac{\sin 13^{\circ}}{\sin 73^{\circ}} \cdot \frac{\sin 77^{\circ}}{\sin 26^{\circ}} \cdot \frac{\sin x}{\sin \left(x+43^{\circ}\right)} .
\end{aligned}
$$

Then $\frac{\sin \left(x+43^{\circ}\right)}{\sin x}=\frac{\sin 13^{\circ} \cdot \sin 77^{\circ}}{\sin 73^{\circ} \cdot \sin 26^{\circ}}$
$$
\begin{array}{l}
=\frac{\sin 13^{\circ} \cdot \cos 13^{\circ}}{\sin 73^{\circ} \cdot \sin 26^{\circ}}=\frac{1}{2 \sin 73^{\circ}}=\frac{\sin 30^{\circ}}{\sin 73^{\circ}} \\
=\frac{\sin 150^{\circ}}{\sin 107^{\circ}}=\frac{\sin \left(107^{\circ}+43^{\circ}\right)}{\sin 107^{\circ}} .
\end{array}
$$

Thus $\cos 43^{\circ}+\cot x \cdot \sin 43^{\circ}$
$$
=\cos 43^{\circ}+\cot 107^{\circ} \cdot \sin 43^{\circ} .
$$

Therefore, $\angle A D B=x=107^{\circ}$.
Solution 2: Let $\angle A D B=x$, then,
$$
\angle D A B=154^{\circ}-x \text {. }
$$

Applying the trigonometric form of Ceva's Theorem to $\triangle D A C$ and point $B$ we have
$$
\begin{array}{l}
1=\frac{\sin \angle A D B}{\sin \angle B D C} \cdot \frac{\sin \angle D C B}{\sin \angle B C A} \cdot \frac{\sin \angle C A B}{\sin \angle B A D} \\
=\frac{\sin x}{\sin 43^{\circ}} \cdot \frac{\sin 86^{\circ}}{\sin 73^{\circ}} \cdot \frac{\sin 30^{\circ}}{\sin \left(154^{\circ}-x\right)} . \\
\text { Then } \frac{\sin \left(154^{\circ}-x\right)}{\sin x}=\frac{\sin 86^{\circ}}{2 \sin 43^{\circ} \cdot \sin 73^{\circ}} \\
=\frac{\cos 43^{\circ}}{\sin 73^{\circ}}=\frac{\sin \left(154^{\circ}-107^{\circ}\right)}{\sin 107^{\circ}} .
\end{array}
$$

Since $\frac{\sin \left(154^{\circ}-x\right)}{\sin x}=\sin 154^{\circ} \cdot \cot x-\cos 154^{\circ}$
is a strictly decreasing function of $x$ in $(0, \pi)$, therefore,
$$
\angle A D B=x=107^{\circ} .
$$