(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