8. In the cyclic quadrilateral $A B C D$, $\angle A C B=15^{\circ}, \angle C A D=45^{\circ}, B C=\sqrt{6}$, $A C=2 \sqrt{3}, A D=$ $\qquad$ .
8. $2 \sqrt{6}$.

In $\triangle A B C$, $A B^{2}=(2 \sqrt{3})^{2}+(\sqrt{6})^{2}-2 \times 2 \sqrt{3} \times \sqrt{6} \cos 15^{\circ}=12-6 \sqrt{3}$, so $A B=3-$ $\sqrt{3} \cdot \frac{\sqrt{6}}{\sin \angle B A C}=\frac{3-\sqrt{3}}{\sin 15^{\circ}} \Rightarrow \sin \angle B A C=\frac{1}{2} \angle B A C+\angle C A D<180^{\circ} \Rightarrow \angle B A C=30^{\circ}$, so $\angle D C A=180^{\circ}-\left(30^{\circ}+45^{\circ}\right)-15^{\circ}=90^{\circ}$. Also, $\angle C A D=45^{\circ}$, hence $A D=2 \sqrt{3} \cdot \sqrt{2}=2 \sqrt{6}$.