设∠CDA=x,∠ABC=y,根据DA=DB=DC=BC,求得x=2y,由四边形的内角和是360°得∠BAC=360°-∠DBA-∠DCA-∠BDC,解得即可得出答案.
解;设∠CDA=x,∠ABC=y,
∵DA=DB=DC=BC,
∴∠BDC=∠DBC=∠DCB=60°,
∠DBA=∠DAB,∠DAC=∠DCA,
∵∠DBA+∠BAD+∠BDA=180°,
∴60°-x+2(60°+y)=180°,
即x=2y,
∠BAC=360°-∠DBA-∠DCA-∠BDC,
=360°-(60°+y)--60°,
=150°.