Answer: D
AP = PC
∴ \(\angle A = \angle PCA\)
Similarly, \(\angle B = \angle PCB\)
\(\angle ACB = \angle PCB + \angle PCA = \angle A + \angle B\)
In \(\triangle ABC\), \(\angle A + \angle B + \angle C = 180°\)
or, \(\angle A + \angle B + \angle A + \angle B = 180°\)
or, \(\angle A + \angle B = 90°\)
∴ \(\angle ACB = \angle A + \angle B = 90°\)
AP = PC
∴ \(\angle A = \angle PCA\)
Similarly, \(\angle B = \angle PCB\)
\(\angle ACB = \angle PCB + \angle PCA = \angle A + \angle B\)
In \(\triangle ABC\), \(\angle A + \angle B + \angle C = 180°\)
or, \(\angle A + \angle B + \angle A + \angle B = 180°\)
or, \(\angle A + \angle B = 90°\)
∴ \(\angle ACB = \angle A + \angle B = 90°\)