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Mathematics angle bisector of the second day of junior high school
Take a point f on the side OF BC, make BF = be, and connect of.

∫BD is the bisector, BF = be, BO is the common edge,

∴△BEO≌△BFO →∠EOB=∠FOB=∠COD

∠∠A = 60

∠ EOB =∠ CBO +∠ BCO, BD and CE are angular bisectors.

∫∠EOB = 1/2( 180-60)= 60

∠ COF = 180-∠ FOB -∠ COD = 60.

And ∵CE are bisectors and CO is a common edge.

∠ COF =∠ Chemical oxygen demand = 60 (certification)

∴△COF≌△COD →CF=CD

∴BC=BF+CF=BE+CD