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證明:∵∠ABC與∠ACB的平分線相交於點O,∴∠OBC=12∠ABC,∠OCB=12∠ACB,∴∠OBC+∠OCB=12(∠ABC+∠ACB),在△OBC中,∠BOC=180°-(∠OBC+∠OCB)=180°-12(∠ABC+∠ACB)=180°-12(180°-∠A)=90°+12∠A…
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