Therefore, in the right triangle PEB, BE= BP= BC=PC,

∴∠BPE=30，

∫∠EPF = 60，

∴FP⊥BC，

∠∠B =∠C = 60，BE=PC，∠PEB=∠FPC=90，

∴△BEP≌△FPC，

∴BE=PF，

∫∠EPF = 60，

△ EPF is an equilateral triangle.

(2) e is EH⊥BC in h,

According to (1), FP⊥BC, FC=BP= BC=4, BE=CP= BC=2,

In triangular FCP, PFC = 90-∠ C = 30,

∫∠PFE = 60，

∴∠GFC=90，

In the right triangle FGC, ∠ c = 60, CF=4,

∴GC=2CF=8，

∴GB=GC-BC=2，

In the right triangle BEP, ∠ EBP = 60, BP=4,

∴PE=2，BE=2，

∴EH=BE？ PE÷BP=，

∴S△GBE= BG？ EH =；

(3)∫CF = 2，AC=6，

∴CF= AC=PC，

△ CPF is an equilateral triangle,

∴∠FPC=60，

∴∠BPE= 180-60-60=60，

∫∠b = 60，

∴△EBP is an equilateral triangle,

∴∠BEP=∠PFC=60，

∴∠PEA=∠PFA，

∠∠A =∠EPF = 60，

∴ Quadrilateral EPFA is a parallelogram,

∴PE=AF=6-2=4.