It is proved that ∵ quadrilateral ABCD is a parallelogram, ∴ AD ∥ BC. ∴ DGC = ∠ GCB,

∵DG=DC,∴∠DGC=∠DCG。 ∴∠DCG=∠GCB。

∠∠DCG+∠DCP = 180，∠GCB+∠FCP= 180 ,∴∠DCP=∠FCP。

∵ In △PCF and △PCE, CE=CF, ∠FCP=∠ECP, CP=CP,

∴△PCF≌△PCE(SAS)。 ∴PF=PE。

Derive ∠DGC=∠GCB from the properties of parallelogram, ∠DGC=∠DCG from the properties of isosceles triangle, ∠DCG=∠GCB, ∠DCP=∠FCP from the equivalence of equilateral complementary angles, and