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

（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

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