: (1) connects BE and PD, the intersection point p is the vertical line of AD, and the vertical foot is g,

① Because point O is the AC midpoint of the diagonal of square ABCD,

∴ Point O is the center of a square, and AC divides∝∠ ∝∠dab and∠∠ ∝∠dcb equally.

∵PE⊥PB,BC⊥CE，

∴B, c, e, p four-point * * * cycle

∴∠PEB=∠PCB=45，∠PBE=∠PCE=45，

∴∠PBE=∠PEB=45，

∴△PBE is an isosceles right triangle,

∴PB=PE，

In △PAB and △PAD, AB=AD, ∠ BAP = ∠ DAP = 45, and AP is the common side.

∴△PAB≌△PAD(SAS)，

∴PB=PD，

∴PE=PD，

And ∵PF⊥CD,

∴df=ef；

②∵PF⊥CD, PG⊥AD and ∠ PCF = ∠ PAG = 45,

∴△PCF and △PAG are isosceles right triangles,

∵ Quadrilateral DFPG is a rectangle,

∴PA= 2PG，PC= 2CF，

PG = DF，DF=EF，

∴PA= 2EF，

∴pc= 2cf = 2(ce+ef)= 2ce+2ef = 2ce+pa，

That is, the relationship among PC, PA and CE is: PC = 2CE+PA；; ;

(2) Conclusion ① is still valid; Conclusion ② It is not valid. At this time, the quantitative relationship of the three line segments in ② is PA-PC = 2ce.

As shown in the figure:

①∵PB⊥PE,BC⊥CE，

∴B, p, c, e four-point * * * cycle

∴∠PEC=∠PBC，

△PBC and△ △PDC include BC=DC (known), ∠ PCB = ∠ PCD = 45 (proved), and PCs have something in common.

∴△PBC≌△PDC(SAS)，

∴∠PBC=∠PDC，

∴∠PEC=∠PDC，

∵PF⊥DE，

∴df=ef；

② Similarly: PA= 2PG= 2DF= 2EF, PC= 2CF,

∴pa= 2ef = 2(ce+cf)= 2ce+2cf = 2ce+PC

That is, the relationship among PC, PA and CE is: PA-PC = 2ce.