(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(jié)DE,DF. ··········································································· 5分
∵△ABC是等邊三角形,∴AB=AC=BC.
又∵D,E,F(xiàn)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(xiàn)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(jié)DF,NF. ······················································································ 5分
∵△ABC是等邊三角形,∴AC=BC=AC.
又∵D,E,F(xiàn)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(gòu)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(jié)論仍然成立(或MF=NE成立). ················ 11分