Question

已知：如图1，点C为线段AB上一点，△ACM，△CBN都是等边三角形，AN交MC于点E，BM交CN于点F．(答好有追问）

已知：如图1，点C为线段AB上一点，△ACM，△CBN都是等边三角形，AN交MC于点E，BM交CN于点F．
(1)CE=CF
(2)EF∥AB
要过程
答好有追加悬赏

Answer 1

∠BAM=∠BCN=60°,∠ACM=∠ABN=60°
=>AM//CN,CM//BN
∠AEM=∠CEN,∠CFM=∠BFN
=>△AEM∽△CEN,△CFM∽△BFN
=>ME:EC=AM:CN,MF:FB=CM:BN
AM=CM,CN=BN
=>AM:CN=CM:BN
=>ME:EC=MF:FB
=>EF//CB (2)
=>∠CFE=∠BCF,∠CEF=∠ACE
=>∠CEF=∠CFE
=>CE=CF (1)

Answer 2

∠BAM=∠BCN=60°,∠ACM=∠ABN=60°
=>AM//CN,CM//BN
∠AEM=∠CEN,∠CFM=∠BFN
=>△AEM∽△CEN,△CFM∽△BFN
=>ME:EC=AM:CN,MF:FB=CM:BN
AM=CM,CN=BN
=>AM:CN=CM:BN
=>ME:EC=MF:FB
=>EF//CB (2)
=>∠CFE=∠BCF,∠CEF=∠ACE
=>∠CEF=∠CFE
=>CE=CF (1)

Answer 3

图呢？

追问

图