证明：连接$BD.$因为$C$是$\overset{\frown}{BD}$的中点，所以$\overset{\frown}{BC}=\overset{\frown}{DC}，$所以$BC = DC.$因为$DC = CP，$所以$BC = DC = CP，$所以$B，$$D，$$P$三点共圆，且$BP$为该圆直径，所以$\angle BDP = 90^{\circ}，$所以$\angle ADB = 180^{\circ}-\angle BDP = 90^{\circ}，$所以$AB$是$\odot O$的直径.
