Qua B vẽ tia \(Bz||Cx\) (1)
Do \(Bz||Cx\) nên \(\widehat{CBz}=\widehat{BCx}\) (hai góc so le trong)
Suy ra \(\widehat{CBz}=50^0\)
Mà \(\widehat{CBz}+\widehat{ABz}=\widehat{CBA}\)
\(50^0+\widehat{ABz}=90^0\)
\(\widehat{ABz}=90^0-50^0=40^0\)
Suy ra \(\widehat{ABz}=\widehat{BAy}\)
Mà \(\widehat{ABz}\) và \(\widehat{BAy}\) là hai góc so le trong
Suy ra \(Bz||Ay\) (2)
Từ (1);(2) suy ra \(Cx||Ay\)