Question

（本小题满分11分）
如图，已知等边三角形ABC中，点D，E，F分别为边AB，AC，BC的中点，M为直线
BC上一动点，△DMN为等边三角形（点M的位置改变时，△DMN也随之整体移动）．
（1）如图①，当点M在点B左侧时，请你判断EN与MF有怎样的数量关系？点F与直线EN有怎样的位置关系？都请直接写出结论，不必证明或说明理由；
（2）如图②，当点M在BC上时，其它条件不变，（1）的结论中EN与MF的数量关系是否仍然成立?若成立，请利用图②证明；若不成立，请说明理由；
（3）若点M在点C右侧时，请你在图③中画出相应的图形，并判断（1）的结论中EN与MF的数量关系及点F与直线EN的位置关系是否仍然成立?若成立?请直接写出结论，不必证明或说明理由．

Answer 1

（1）判断：EN与MF相等（或EN=MF），点F在直线NE上   ······ 3分
（说明：答对一个给2分）
（2）成立．································ 4分
证明：
法一：连结DE，DF．   ··········································································· 5分
∵△ABC是等边三角形，∴AB=AC=BC．
又∵D，E，F是三边的中点，
∴DE，DF，EF为三角形的中位线．∴DE=DF=EF，∠FDE=60°．
又∠MDF+∠FDN=60°，∠NDE+∠FDN=60°，
∴∠MDF=∠NDE． ················································································ 7分
在△DMF和△DNE中，DF=DE，DM=DN，∠MDF=∠NDE，
∴△DMF≌△DNE． ··············································································· 8分
∴MF=NE．      　··············································································· 9分

法二：
延长EN，则EN过点F．    ······································································ 5分
∵△ABC是等边三角形，∴AB=AC=BC．又∵D，E，F是三边的中点，∴EF=DF=BF．  
∵∠BDM+∠MDF=60°，∠FDN+∠MDF=60°，∴∠BDM=∠FDN．······················· 7分
又∵DM=DN，∠ABM=∠DFN=60°，∴△DBM≌△DFN．································· 8分
∴BM=FN．∵BF=EF， ∴MF=EN．···························································· 9分
法三：
连结DF，NF． ······················································································ 5分

∵△ABC是等边三角形，∴AC=BC=AC．
又∵D，E，F是三边的中点，∴DF为三角形的中位线，∴DF=AC=AB=DB．
又∠BDM+∠MDF=60°，∠NDF+∠MDF=60°，∴∠BDM=∠FDN． ………………7分
在△DBM和△DFN中，DF=DB，
DM=DN，∠BDM=∠NDF，∴△DBM≌△DFN．
∴∠B=∠DFN=60°．…………………………………………………………………8分
又∵△DEF是△ABC各边中点所构成的三角形，
∴∠DFE=60°．∴可得点N在EF上，∴MF=EN．………………………………9分
（3）画出图形（连出线段NE）， ······························································· 10分
MF与EN相等及点F在直线NE上的结论仍然成立（或MF=NE成立）． ················ 11分

略