Question

（本小題滿分11分）
如圖，已知等邊三角形ABC中，點D，E，F分別為邊AB，AC，BC的中點，M為直線
BC上一動點，△DMN為等邊三角形（點M的位置改變時，△DMN也隨之整體移動）．
（1）如圖①，當點M在點B左側時，請你判斷EN與MF有怎樣的數量關系？點F與直線EN有怎樣的位置關系？都請直接寫出結論，不必證明或說明理由；
（2）如圖②，當點M在BC上時，其它條件不變，（1）的結論中EN與MF的數量關系是否仍然成立?若成立，請利用圖②證明；若不成立，請說明理由；
（3）若點M在點C右側時，請你在圖③中畫出相應的圖形，并判斷（1）的結論中EN與MF的數量關系及點F與直線EN的位置關系是否仍然成立?若成立?請直接寫出結論，不必證明或說明理由．

Answer 1

（1）判斷：EN與MF相等（或EN=MF），點F在直線NE上   ······ 3分
（說明：答對一個給2分）
（2）成立．································ 4分
證明：
法一：連結DE，DF．   ··········································································· 5分
∵△ABC是等邊三角形，∴AB=AC=BC．
又∵D，E，F是三邊的中點，
∴DE，DF，EF為三角形的中位線．∴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：
延長EN，則EN過點F．    ······································································ 5分
∵△ABC是等邊三角形，∴AB=AC=BC．又∵D，E，F是三邊的中點，∴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：
連結DF，NF． ······················································································ 5分

∵△ABC是等邊三角形，∴AC=BC=AC．
又∵D，E，F是三邊的中點，∴DF為三角形的中位線，∴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各邊中點所構成的三角形，
∴∠DFE=60°．∴可得點N在EF上，∴MF=EN．………………………………9分
（3）畫出圖形（連出線段NE）， ······························································· 10分
MF與EN相等及點F在直線NE上的結論仍然成立（或MF=NE成立）． ················ 11分

解析