（1）证明：$\because AB$是半圆$O$的直径，$\therefore\angle ACB = 90^{\circ}。$$\because CP$是半圆$O$的切线，$\therefore\angle OCP = 90^{\circ}。$$\therefore\angle ACB=\angle OCP。$$\therefore\angle ACB-\angle OCB=\angle OCP-\angle OCB。$$\therefore\angle ACO=\angle BCP。$
（2）解：由（1），知$\angle ACO=\angle BCP。$$\because\angle ABC = 2\angle BCP，$$\therefore\angle ABC = 2\angle ACO。$$\because OA = OC，$$\therefore\angle ACO=\angle BAC。$$\therefore\angle ABC = 2\angle BAC。$$\because\angle ABC+\angle BAC = 90^{\circ}，$$\therefore\angle BAC = 30^{\circ}，$$\angle ABC = 60^{\circ}。$$\therefore\angle ACO=\angle BCP = 30^{\circ}。$$\therefore\angle P=\angle ABC-\angle BCP = 60^{\circ}-30^{\circ}=30^{\circ}。$
（3）解：由（2），知$\angle BAC = 30^{\circ}。$$\because\angle ACB = 90^{\circ}，$$\therefore BC=\frac{1}{2}AB = 2，$则$AC=\sqrt{AB^{2}-BC^{2}}=2\sqrt{3}。$$\therefore S_{\triangle ABC}=\frac{1}{2}BC\cdot AC=\frac{1}{2}\times2\times2\sqrt{3}=2\sqrt{3}。$$\therefore$阴影部分的面积是$\frac{1}{2}\pi\times(\frac{AB}{2})^{2}-2\sqrt{3}=2\pi-2\sqrt{3}。$
