解：$\because AB$是$\odot O$的直径，$\therefore \angle ACB = \angle ADB = 90^{\circ}.$ 在$Rt\triangle ABC$中，$\because AB = 6，$$AC = 2，$$\therefore BC = \sqrt{AB^{2}-AC^{2}}=\sqrt{6^{2}-2^{2}} = 4\sqrt{2}.$$\because \angle ACB$的平分线交$\odot O$于点$D，$$\therefore \angle DCA=\angle BCD.$$\therefore AD = BD.$$\therefore$在$Rt\triangle ABD$中，易得$AD = BD = \frac{\sqrt{2}}{2}AB = 3\sqrt{2}.$$\therefore S_{四边形ADBC}=S_{\triangle ABC}+S_{\triangle ABD}=\frac{1}{2}AC\cdot BC+\frac{1}{2}AD\cdot BD=\frac{1}{2}\times2\times4\sqrt{2}+\frac{1}{2}\times(3\sqrt{2})^{2}=4\sqrt{2}+9$
