Question

如图1.点C为线段AB上一点,△ACM,△CBN是等边三角形,直线AN,MC交于点E,直线BM.CN交于点F.（1）求证：AN=BM;

（2）求证△CEF是等边三角形，（3）将△ACM绕点C按逆时针方向旋转60度,其他条件不变,在图2中补出符合要求的图形,并判断第(1) (2)两小提结论是否仍然成立.

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)