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.