Advertising split ∠BAC

∴ ∠ 1=∠2

∫BI average score ∠ABC

∴ ∠3=∠4

∠∠2 and∠∠∠∠ 5 are the circumferential angles of the circular arc CD.

∴∠2=∠5

∵ ∠ 1+∠3=∠6,∠3=∠4,∠ 1=∠2,∠2=∠5

∴ ∠4+∠5=∠6

∴ ∠DBI=∠DIB

∴ BD=ID

∴ ∠ 1=2

∴ BD=CD

∴BD=DC=DI

⑵ o For OE⊥BC in E, connect OC.

∵∠BAC= 120

∴ ∠BDC=60

BD = CD

Delta BDC is an equilateral triangle.

∠∠BDC = 60

∴∠COE=60

∫OC = 10，∠COE=60

∴ CE=OC？ sin60

∴ CE=5√3

∴ BC = 10√3

∴ S△BDC=34×( 10√3)2=75√3