
做正三角形ADE、CDF,连接AF、EF
过E点做AD垂线EH与AC相交于G,连接GD、GF
∠ACF=∠ACD=360-(180-6-24)-60=150°
AC共边,CD=CF
∴△ACD≌△ACF
∵AD=ED,CD=DF,∠ADC=∠EDF=60-∠BDE=24°
∴△ACD≌△EDF
∵AH=HD
∴AG=GD=GF
∠GFA=∠GDA=∠GAD=6°
AE=AF
∠AFE=∠AEF=60+6=66°
∠GFE=∠AFE+∠GFA=66+6=72°
∠EGF=180-∠GFE-∠GEF=180-72-36=72°
∴△EFG是等腰三角形
EG=EF=AC
AE=BC
∠ACB=∠AEG=30°
∴△ABC≌△AEG
∠ABC=∠AGE=60-6=54°