在CB上取点E，使CE=CA，连结AE，又∵∠ACB=60°，∴△ACE是正△，∴∠AEC=60°，∵△ACD≌△ECD∴AD=DE=BE，设∠DAE=X，则∠DEA=X，∠BDE=2X，∠DBE=2X，∠DEC=4X，即∠AEC+∠AED=60+X=4X，解得X=20°∴∠BAC=60°+20°=80°
