Question

Answer 1

Qua B ta vẽ \(Bz\) // \(Ax\)

\(\Leftrightarrow\widehat{xAB}=\widehat{ABz}\) (2 góc so le trong)

Mà \(\widehat{xAB}=60^0\)

\(\Leftrightarrow\widehat{ABz}=60^0\)

Ta có :

\(\widehat{ABz}+\widehat{zBC}=\widehat{ABC}\)

\(\Leftrightarrow\widehat{zBC}=\widehat{ABC}-\widehat{ABz}\)

\(\Leftrightarrow\widehat{zBC}=130^0-60^0\)

\(\Leftrightarrow\widehat{zBC}=70^0\)

Lại có : \(\widehat{BCy}=70^0\)

Mà 2 góc là 2 góc so le trong

\(\Leftrightarrow Bz\) // \(Cy\left(d.h.n.b\right)\)

Ta có :

\(Ax\) // \(Bz\)

\(Cy\) // \(Bz\)

\(\Leftrightarrow Ax\) // \(Cy\)