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證明:作OE⊥AB於E,OF⊥AC於F,∵AO平分∠BAC,∴OE=OF(角平分線上的點到角兩邊的距離相等).∵∠1=∠2,∴OB=OC.∴Rt△OBE≌Rt△OCF(HL).∴∠5=∠6.∴∠1+∠5=∠2+∠6.即∠ABC=∠ACB.∴AB=AC.∴△ABC是等腰三角形.
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