(1) According to the meaning of the question, Xiao Cong's thinking is as follows: By judging that the triangles DHP and PGF are congruent triangles, it is concluded that the triangle HCG is an isosceles triangle and P is the midpoint of the bottom;

(2) The idea is the same as above. Extend the intersection of GP and AD at h point to connect CH and CG. This question not only proves △ GFP △ HDP (P is the midpoint of HG), but also proves △ HDC △ GBC (triangle CHG is an isosceles triangle).

(3) ∠ ABC = ∠ BEF = 2α (0 < α < 90), then ∠ PCG = 90-α, from (1): PG: PC = Tan (90-α).

Solution: (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=tan60 =

three

The positional relationship between pg and PC is PG⊥PC,

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=

three

(2) Conjecture: The conclusion in (1) has not changed.

Proof: as shown in figure 2, extend GP intersection AD to point h and connect CH,

∵P is the midpoint of the straight line DF,

∴FP=DP，

∫AD∨GF，

∴∠HDP=∠GFP，

∠∠GPF =∠HPD，

∴△GFP≌△HDP(ASA)，

∴GP=HP,GF=HD，

The quadrilateral ABCD is a diamond,

∴CD=CB,∠HDC=∠ABC=60，

∵∠ ABC =∠ BEF = 60, the diagonal BF of the rhombic BEFG is just in a straight line with the side AB of the rhombic ABCD.

∴∠GBF=60，

∴∠HDC=∠GBF，

∵ quadrilateral BEFG is a diamond,

∴GF=GB，

∴HD=GB，

∴△HDC≌△GBC，

∴CH=CG,∠HCD=∠GCB

∴PG⊥PC (the point with equal distance to both ends of the line segment is on the middle vertical line of the line segment)

∫∠ABC = 60

∴∠DCB=∠HCD+∠HCB= 120

∠HCG=∠HCB+∠GCB

∴∠HCG= 120

∴∠GCP=60

∴

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=tan∠GCP=tan60 =

three

(3)∫∠ABC =∠BEF = 2α(0