∵ quadrilateral ABCD is a square,

∴AC bisection ∠BCD, CB=CD,

∴△BCP≌△DCP.

∴∠PBC=∠PDC,PB=PD.

∵PB⊥PE,∠BCD=90，

∴∠PBC+∠PEC=

360 -∠BPE-∠BCE= 180

∴∠PED=∠PBC=∠PDC.∴PD=PE.

∵PF⊥CD,∴DF=EF.

②PC-PA=■CE。

The proof is as follows: the intersection p is the PH⊥AD of point H.

From ①: PA=■PH=■DF=■EF, PC =■ cf.

∴PC-PA=■(CF-EF).

That is to say, PC-PA.

Method 2:

① It is proved that the intersection point P is GH⊥AD at H point and BC at G point.

∫AC is the diagonal of square ABCD, while PF⊥CD.

∴GB=HA=HP=DF.

∵PB⊥PE,∴∠GPB+∠GPE=90。

∠∠GPE+∠EPF = 90，

∴∠EPF=∠BPG.

∠∠pfe =∠pgb = 90，

∴△PEF≌△PBG.

∴BG=EF.∴DF=EF.

②PC-PA=■CE。

The proof is as follows: pass point E as ET⊥HG and give it to PC at point K.

From ①: HP=DF=EF=PT. CK=■CE。

∠∠APH =∠KPT = 45，∠ AHP =∠ KTP = 90。

∴△PKT≌△PAH.∴PA=PK.

∴PC-PA=PC-PK=CK=■CE.

(2) solution: drawing;

Conclusion ① It is still valid;

Conclusion ② Not true,

At this time, the three lines in ③

The quantitative relationship between segments is

PA-PC=■CE。