Solution: (1)① Extend the intersection of FP and AB to G because PF∨AD∨BC∠PCF =∠FPC = 45, then GB=PF=FC.

Because Pb ⊥ PE ∠ GPB+∠ FPE = 90 ∠ GBP = ∠ FPE ∠ PGB = ∠ PFE = 90.

So △PGB=△EFP, so EF=GP=GA=DF.

②CE=CF-EF=PC/ radical number two -PA/ radical number two, so PC=PA+ radical number two CE.

(2)DF-EF is still valid. The same method as above is still full of certificates.

PA=PC+ root sign CE