答案：1. C 点拨:$\because △ ABC$与$△ CDE$都是等边三角形,$\therefore AC=BC,EC=DC,∠ ACB = ∠ DCE = 60°, \therefore ∠ ACD = ∠ BCE=120°, \therefore △ ACD ≌ △ BCE(\mathrm{SAS}), \therefore AD=BE, ∠ CAD=∠ CBE. \because BM=\frac{1}{3}BE,AN=\frac{1}{3}AD, \therefore BM=AN, \therefore △ BCM ≌ △ ACN(\mathrm{SAS}), \therefore CM=CN, ∠ BCM=∠ ACN, \therefore ∠ MCN = ∠ ACM + ∠ ACN = ∠ ACM + ∠ BCM = ∠ ACB=60°, \therefore △ CMN$是等边三角形.

答案：
2. 解:$BM=AM+CM$. 理由:如答图,在 $DA$ 上取点 $F$,使$DF=ME$,连接 $CF$.

$\because △ ABC$与$△ CDE$都是等边三角形，
$\therefore AC=BC,EC=DC,∠ ACB=∠ DCE=60°$,
$\therefore ∠ ACD=∠ BCE=120°, \therefore △ ACD ≌ △ BCE(\mathrm{SAS})$,
$\therefore AD=BE,∠ CAD=∠ CBE$.
$\because DF=ME, \therefore AD-DF=BE-ME$,即 $AF=BM$.
同理可证$△ MCE ≌ △ FCD, \therefore CM=CF,∠ DCF=∠ ECM$,
$\therefore ∠ MCF = ∠ MCE + ∠ ECF = ∠ ECF + ∠ FCD = ∠ ECD=60°, \therefore △ MCF$ 是等边三角形, $\therefore CM=MF$.
$\because AF=AM+MF=AM+CM, \therefore BM=AM+CM$.