It is proved that P2 is the bisector of AB, so BP2/BA = BP2/CD = BO3/DO3 =1/2.

So BP3/BA=BO3/BD= 1/3, P3 is the bisector of AB.

bo4/Do4 = BP3/CD = BP3/BA = 1/3，

BP4/BA=BO4/BD= 1/4，

So P4 is the rendezvous point of AB,

Similar to provABle BPn/BA=BOn/BD= 1/n, so Pn is the n-bisector of ab.

(pay attention to the subscript, it's not easy to type)