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.