1、∵AB=AC∴∠B=∠C∵EF∥BC∴∠AEF=∠B,∠AFE=∠C∴∠AEF=∠AFE∵AD⊥BC∴由等腰三角形大家平分线,底边的高,中线三线合一得:∠BAD=∠CAD即∠EAG=∠FAG∵AG=AG∴△AEG≌△AFG(AAS)∴EG=FG∠AGE=∠AGF∵∠AGE+∠AGF=180°∴∠AGE=∠AGF=90°即AG⊥EF∴AD垂直平分EF2、△AGE和△AGF△DEG和△DFG△AED和△AFD△ABD和△ACD△BDE和△CDF