Answer: D
AP = PC
∴\(\angle\)PAC = \(\angle\)PCA
Similarly, \(\angle\)PBC = \(\angle\)PCB
\(\angle\)ACB = \(\angle\)PCB + \(\angle\)PCA = \(\angle\)PAC + \(\angle\)PBC
In ∆ABC,
\(\angle\)BAC + \(\angle\)ABC + \(\angle\)BCA = 180°
Or, \(\angle\)PAC + \(\angle\)PBC + \(\angle\)PAC + \(\angle\)PBC = 180°
Thus, \(\angle\)PAC + \(\angle\)PBC = 90°
∴\(\angle\)ACB = \(\angle\)PAC + \(\angle\)PBC = 90°
AP = PC
∴\(\angle\)PAC = \(\angle\)PCA
Similarly, \(\angle\)PBC = \(\angle\)PCB
\(\angle\)ACB = \(\angle\)PCB + \(\angle\)PCA = \(\angle\)PAC + \(\angle\)PBC
In ∆ABC,
\(\angle\)BAC + \(\angle\)ABC + \(\angle\)BCA = 180°
Or, \(\angle\)PAC + \(\angle\)PBC + \(\angle\)PAC + \(\angle\)PBC = 180°
Thus, \(\angle\)PAC + \(\angle\)PBC = 90°
∴\(\angle\)ACB = \(\angle\)PAC + \(\angle\)PBC = 90°