Question

第20题怎么做

Answer 1

延长AD,取AF=AB,连接BF,EF,,取EG=AE,连接AG
∵AB=AC
∴∠ABD=∠ACD
∵AB=AF,∠BAD=60 ,
∴△ABF是等边三角形
∴AB=AF
∵AD是∠DAC平分线
∴∠EAF=∠CAE
∵AF=AB=AC,∠EAF=∠CAE,AE=AE
∴△AEF≌△ACE(SAS)

∴∠AFE=∠ACD,EF=EC
∵∠ABD=∠ACD
∴∠ABD=∠DFE
又∠ABD=∠AFE,∠ADB=∠EDF
∴∠DEF=∠BAD=60
∴△ABD∽△DEF

∴DB/DF=AD/DE
∵DB/DF=AD/DE,∠BDF=∠ADE
∴△BDF∽△ADE
∴∠AED=∠AFB=60
∵EG=AE,∠AFB=60
∴△AEG是等边三角形
∴EG=AE=AG,
∵AB=AC,∠BAG=∠CAE,AG=AE
∴△ABG≌△ACE(SAS)
∴BG=CE=3
∵BE=BD+DE=BG+EG=,BD-AE=1 ,BG=CE=3 ,EG=AE
∴DE=BG-(BD-AE)
∴DE=2
∵△DEF∽△ABD
∴DE/DF=EF/AB ,DE=3,EF=3
∴AB/AD==3/2
∴DF=AB-AD=3/2*AD-AD=1/2*AD
∴DF/AD=1/2
∵∠AGD=∠DEF=60
∴EF//AG
∴DE/DG=DF/AD=1/2
∴DG=2DE=4
∴BD=BG+DG=3+4=7
又△ADG∽△ABD
∴AD^2=BD*DG
∴AD^2=7*4=28
∴AD=2√7