Let ED=x, then EP=AE=2-x,
In Rt△EDP, EP2=ED2+DP2,
Namely, (2-x)2=x2+ 12,
Solution: x=34,
∠∠PED+∠EPD = 180-∠D = 180-90 = 90,
∠EPD+∠GPC = 180-∠EPG = 180-90 = 90,
∴∠EPD=∠GPC,
∠∠d =∠c = 90,
∴△EPD∽△PGC,
∴△EDP and△△△ PCG perimeter ratio =DEPC=34,
∴ Let C 1 be the perimeter of △PCG and C2 be the perimeter of △PDE, then C 1: C2 = 4: 3.
So the answer is: 4: 3.