本题只要证明GM//BE即可得证。设∠CBE=∠EBF=∠FBA=a,AB=AC,∠ACB= ∠ABC=3a 连接GM，ME，∵AD是等腰三角形ABC的底边上BC的高,∴NB=NC,∠NCB=∠CBE=a,∠GNB=∠NCB+∠CBE=2a, ∵∠BAD=∠CAD=1/2∠BAC,∠EBF=∠FBA=a,∴M是ΔABE的内心,∴∵∠AEB=∠AEM+∠BEM=2∠BEM,∠AEB=∠ACB+ ∠CBE=4a, ∴ 2∠BEM=4a ,∠BEM=2a ,∵ΔBMD≌ΔCMD,∠BND=∠CND,NB...
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