Giải :
Cho tam giác \(\widehat{ABC}\) và gọi các góc ngoài của tam \(\widehat{ABC}\) lần lượt là : \(\widehat{A'}\) ,\(\widehat{B'}\) ,\(\widehat{C'}\)
Ta có : \(\widehat{A}\) + \(\widehat{B}\) + \(\widehat{C}\) = \(180^0\)
Mà : \(\widehat{A'}\) = \(\widehat{B}\) + \(\widehat{C}\)
\(\widehat{B'}\) = \(\widehat{A}\) + \(\widehat{C}\)
\(\widehat{C'}\) = \(\widehat{A}\) + \(\widehat{B}\)
=> \(\widehat{A'}\) + \(\widehat{B'}\) + \(\widehat{C'}\) = 2\(\widehat{A}\) + 2\(\widehat{B}\) + 2\(\widehat{C}\)
=> \(\widehat{A'}\) + \(\widehat{B'}\) + \(\widehat{C'}\) = 2 . (\(\widehat{A}\) + \(\widehat{B}\) + \(\widehat{C}\))
=> \(\widehat{A'}\) + \(\widehat{B'}\) + \(\widehat{C'}\) = 2. \(180^0\)
=> \(\widehat{A'}\) + \(\widehat{B'}\) + \(\widehat{C'}\) = \(360^0\)
Vậy tổng 3 góc ngoài của 1 tam giác là : \(360^0\)
