Please read the following materials: problem: as shown in Figure 1, in diamond ABCD and diamond befg, points a, B and E are on the same line, P is the midpoint of line DF, connecting PG and PC. if ∠ ABC = ∠ bef = 60 °, explore the position relationship between PG and PC and the value of pgpc. XiaoCong's idea is: extend GP intersection DC at point h, construct congruent triangle, and solve the problem through reasoning. Please refer to XiaoCong Students' ideas, explore and solve the following problems: (1) write the position relationship between PG and PC and the value of pgpc in the above problem; (2) rotate the diamond BFG in Figure 1 clockwise around point B, so that the diagonal BF of diamond BFG is in the same line with the edge ab of diamond ABCD, and other conditions in the original problem remain unchanged (as shown in Figure 2). Are the two conclusions you get in (1) correct What's the change? Write down your conjecture and prove it. (3) if ∠ ABC = ∠ bef = 2 α (0 °＜ α＜ 90 °) in Figure 1, rotate the diamond BFG clockwise at any angle around point B, and other conditions in the original problem remain unchanged, please write down the value of pgpc directly (expressed by the formula containing α)

Please read the following materials: problem: as shown in Figure 1, in diamond ABCD and diamond befg, points a, B and E are on the same line, P is the midpoint of line DF, connecting PG and PC. if ∠ ABC = ∠ bef = 60 °, explore the position relationship between PG and PC and the value of pgpc. XiaoCong's idea is: extend GP intersection DC at point h, construct congruent triangle, and solve the problem through reasoning. Please refer to XiaoCong Students' ideas, explore and solve the following problems: (1) write the position relationship between PG and PC and the value of pgpc in the above problem; (2) rotate the diamond BFG in Figure 1 clockwise around point B, so that the diagonal BF of diamond BFG is in the same line with the edge ab of diamond ABCD, and other conditions in the original problem remain unchanged (as shown in Figure 2). Are the two conclusions you get in (1) correct What's the change? Write down your conjecture and prove it. (3) if ∠ ABC = ∠ bef = 2 α (0 °＜ α＜ 90 °) in Figure 1, rotate the diamond BFG clockwise at any angle around point B, and other conditions in the original problem remain unchanged, please write down the value of pgpc directly (expressed by the formula containing α)

(1) ∵ CD ∥ GF, ∠ PDH = ∠ PFG, ∠ DHP = ∠ PGF, DP = PF, ≌ DPH ≌ FGP, ≌ pH = PG, DH = GF, ∵ CD = BC, GF = GB = DH, ≁ ch = CG, ≁ CP ⊥ Hg, ≁ ABC = 60 °, ≁ DCG = 120 °, ≌ PCG = 60 °, ≌ PG: PC = Tan 60 ° = 3, ≁ line PG is related to the position of PC