Kẻ \(Bz//Ax//Cy\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{A}+\widehat{ABz}=180^0\left(trong.cùng.phía\right)\\\widehat{C}+\widehat{CBz}=180^0\left(trong.cùng.phía\right)\end{matrix}\right.\)
Mà \(\widehat{B}=\widehat{ABz}+\widehat{CBz}\)
Vậy \(\widehat{A}+\widehat{B}+\widehat{C}=360^0\)
Kẻ Bz//Ax//Cy
Ta có: Bz//Ax
\(\Rightarrow\widehat{BAx}+\widehat{ABz}=180^0\left(1\right)\)(trong cùng phía)
Ta có: Bz//Cy
\(\Rightarrow\widehat{CBz}+\widehat{BCy}=180^0\left(2\right)\)(trong cùng phía)
\(\left(1\right),\left(2\right)\Rightarrow\widehat{BAx}+\widehat{ABz}+\widehat{CBz}+\widehat{BCy}=180^0+180^0\)
\(\Rightarrow\widehat{A}+\widehat{B}+\widehat{C}=360^0\)
Kẻ Bz // Ax
=>\(\widehat{xAB}+\widehat{ABz}=180^o\left(hai.gocóc.trong.cunùng.phiaía.buù.nhau\right)\)
Bz// Ax mà Ax// Cy => Bz//Cy
=> \(\widehat{zBC}+\widehat{BCy}=180^o\left(hai.gocóc.trong.cungùng.phiaía.buù.nhau\right)\)
Mà \(\widehat{ABz}+\widehat{zBC}=\widehat{B}\)
=> \(\widehat{A}+\widehat{B}+\widehat{C}=180+180=360^o\)
