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.

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:BNAM=CM,CN=BN=>AM:CN=CM:BN=>ME:EC=MF:FB=>EF//CB (2)=>∠CFE=∠BCF,∠CEF=∠ACE=>...