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iceghost

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Qua A vẽ d // BC

Ta có : A^1+A^2+A^3=180o\hat{A}_1 + \hat{A}_2 + \hat{A}_3 = 180^o ( kề bù )
Mà A^2=B^1\hat{A}_2=\hat{B}_1 ( so le trong, d // BC )
A^3=C^1\hat{A}_3 = \hat{C}_1 (so le trong, d // BC )
  ⟹  \implies A^1+B^1+C^1=180o\hat{A}_1+\hat{B}_1+\hat{C}_1=180^o
Vậy tổng ba góc trong △\triangle bằng 180o180^o
 
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