（1）证明：因为$C$为$\overset{\frown}{AB}$的中点，所以$\overset{\frown}{AC} = \overset{\frown}{BC}，$所以$\angle AOC = \angle BOC.$
因为$OA = OB，$所以$OC \perp AB.$
因为$CD// AB，$所以$OC \perp CD.$
因为$OC$是$\odot O$的半径，所以$CD$是$\odot O$的切线.
（2）解：因为$OC \perp CD，$所以$\angle OCD = 90^{\circ}.$
因为$OB = OC = OA = 3，$$BD = 2，$所以$OD = OB + BD = 5，$
根据勾股定理$CD = \sqrt{OD^{2} - OC^{2}} = \sqrt{5^{2} - 3^{2}} = 4，$
所以$S_{\triangle OCD} = \frac{1}{2}CD\cdot OC = \frac{1}{2}×4×3 = 6.$
故$\triangle OCD$的面积为$6.$
