(1)证明:∵AB=AC,∴∠ABC=∠BCA=∠ADB,∵四边形ABCD是圆内接四边形,∴∠CDE=∠ABC,∴∠ADB=∠CDE;(2)解:作AM⊥CD于点M,∵AB=10,AF=6,∴BF=8,∵AD平分∠BDM,AM=AF=6,∴△ACM≌△ABF,∴CM=BF=8,∴DF=DM=CM-CD=2.∴BD=BF+DF=10=AB.∴∠BAD=∠ADB=∠ADM,∴S△DBC=S△ADC= 1 2 CD×AM=18.