2. Pass a point P outside the plane α where △ABC is located, so that it is PO⊥α, and the vertical foot is O, connecting PA, PB and PC.

(1) If PA = Pb = PC and ∠ C = 90, then the O point is the _ _ _ _ point on the AB side.

(2) If PA = Pb = PC, then point O is the _ _ outer _ _ center of △ABC.

(3) If PA⊥PB, PB⊥PC, PC⊥PA, then point O is the _ _ _ _ center of △ABC.

Prove:

①

Connect OA, OB, OC

Pa = Pb = PC, and PO is the common side.

∴Rt△AOP≌Rt△BOP≌Rt△COP

∴OA=OB=OC

∴O is the external center of △ABC.

The answers to (1) and (2) are proved.

②

Connect AO and CO and extend the intersection of BC and AB at point d and point e.

∵PA⊥PC,PB⊥PC

∴ PC⊥ Pabu

∴PC⊥AB

∵PO⊥α

∴PO⊥AB,PO∩PC=P

∴AB⊥CO

Similarly, BC⊥AO

∴O is the intersection of heights.

∴O is the focus of ABC.