Answer: Proof: As shown in the figure, point O in D is OD⊥AB, and point O in E is OE⊥AC.

∵OD⊥AB，OE⊥AC，AO ∠BAC，

∴OD=OE.？

∵∠ 1=∠2,

∴OB=OC.？

Rt△BDO and Rt△CEO

OD=OEOB=OC，

∴△DOB≌△EOC(HL)，

∴∠DBO=∠ECO，

∴∠ABC=∠ACB，

∴AB=AC.

∫OA segmentation ∠BAC,

∴AO⊥BC.