Question

chứng minh tổng ba góc ngoài của tam giác bằng 360

Answer 1

Answer 2

Ta có hình vẽ :

=> \(\left\{{}\begin{matrix}\widehat{x}=\widehat{A}+\widehat{B}\\\widehat{y}=\widehat{B}+\widehat{C}\\\widehat{z}=\widehat{C}+\widehat{A}\end{matrix}\right.\)

=> \(\widehat{x}+\widehat{y}+\widehat{z}=\widehat{A}+\widehat{B}+\widehat{B}+\widehat{C}+\widehat{C}+\widehat{A}\)

= \(\left(\widehat{A}+\widehat{A}\right)+\left(\widehat{B}+\widehat{B}\right)+\left(\widehat{C}+\widehat{C}\right)\)

= \(2\widehat{A}+2\widehat{B}+2\widehat{C}\)

= \(2\left(\widehat{A}+\widehat{B}+\widehat{C}\right)\)

Ta có : \(\widehat{A}+\widehat{B}+\widehat{C}\) = 180^o ( Định lý tổng 3 góc của 1 tam giác )

\(\Rightarrow2\left(\widehat{A}+\widehat{B}+\widehat{C}\right)\) = 2.180^o = 360^o

=> đpcm