证明: (1)过A做MN‖BC 则∠MAB=∠B, ∠NAC=∠C 即∠BAC+∠ABC+∠ACB=∠A+∠MAB+∠NAC 因MN是过A的曲线,所以 ∠A+∠MAB+∠NAC=180° 所以∠BAC+∠ABC+∠ACB=180°
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证明: (1)过A做MN‖BC 则∠MAB=∠B, ∠NAC=∠C 即∠BAC+∠ABC+∠ACB=∠A+∠MAB+∠NAC 因MN是过A的曲线,所以 ∠A+∠MAB+∠NAC=180° 所以∠BAC+∠ABC+∠ACB=180°