10CT2 ** In the convex quadrilateral $ABCD$, $AC$ and $BD$ intersect at point $P, \angle DBC=60^{\circ}$, $\angle ACB=50^{\circ}, \angle ABD=20^{\circ}, \angle ACD=30^{\circ}$, find $\angle ADB$.
Parse 1 $\frac{A P}{A C}=\frac{S_{\triangle A B P}}{S_{\triangle A B C}}=\frac{B P \sin 20^{\circ}}{B C \sin 80^{\circ}}=\frac{\sin 20^{\circ} \sin 50^{\circ}}{\sin 70^{\circ} \sin 80^{\circ}}$,
Ifij $\frac{D P}{C D}=\frac{\sin 30^{\circ}}{\sin 70^{\circ}}$, since $\frac{\sin 20^{\circ} \sin 50^{\circ} \sin 70^{\circ}}{\sin 80^{\circ}}=$
$\frac{2 \cos 10^{\circ} \sin 10^{\circ} \sin 50^{\circ} \sin 70^{\circ}}{\sin 80^{\circ}}-\frac{2}{4} \sin 30^{\circ}-\frac{1}{4}$, thus $\frac{A P}{A C}-$ $\frac{\sin 20^{\circ} \sin 50^{\circ}}{\sin 70^{\circ} \sin 80^{\circ}}=\left(\frac{\sin 30^{\circ}}{\sin 70^{\circ}}\right)^{2}=\frac{D P^{2}}{C D^{2}}$

Now construct $P M / / C D$, with point $M$ on $A D$. Thus $\frac{A P}{A C}=\frac{M P}{C D}$, which means $D P^{2}=$ $M P \cdot C D$, and $\angle M P D=\angle P D C$, so $\triangle M P D \sim \triangle P D C$, hence $\angle A D B=\angle P C D=30^{\circ}$.

Parse 2 Let $\angle A D B=\theta$, then $\frac{\sin \left(\theta+20^{\circ}\right)}{\sin \theta}=\frac{B D}{\Lambda B}=\frac{B D}{B C} \cdot \frac{B C}{\Lambda B}=\frac{\sin 80^{\circ}}{\sin 40^{\circ}}$, which means $\cos 20^{\circ}+\cot \theta \sin 20^{\circ}=2 \cos 40^{\circ}, \cot \theta=\frac{2 \cos 40^{\circ}-\cos 20^{\circ}}{\sin 20^{\circ}}=\frac{2 \cos \left(60^{\circ}-20^{\circ}\right)-\cos 20^{\circ}}{\sin 20^{\circ}}$ $=\frac{2 \sin 60^{\circ} \sin 20^{\circ}}{\sin 20^{\circ}}=\sqrt{3}$. Therefore, $\theta=30^{\circ}$.