2.

three

Prove: connecting OB, OM⊥AB with intersection O of m, ON⊥BC of n.

∫∠ABC = 60

∴∠bac+∠acb= 180-∠abc= 120

∵AD bisection ∠BAC, CE bisection ∠ACB

∴∠OAC=∠BAC/2，∠OCA=∠ACB/2

∴∠aoc= 180-(∠oac+∠oca)= 180-(∠bac+∠acb)/2= 120

∴∠DOE=∠AOC= 120

∴∠ABC+∠DOE= 180

∠∠od b+∠OEB+∠ABC+∠DOE = 180

∴∠ODB+∠OEB= 180

∠∠OEB+∠OEA = 180

∴∠OEA=∠ODB

There are also ∵AD split ∠BAC, CE split ∠ACB.

∴O is the intersection of the bisector of the angle △ABC.

∴OB split equally ∠ABC

∵OM⊥AB,ON⊥BC

∴OM=ON,∠OME=∠OND=90

∴△OME≌△OND？ (AAS)

∴OE=OD

Answer hard. That's it.