a) Vì AB//MN (gt)
=>\(\widehat{ABM}=\widehat{BMN}\) (cặp góc soletrong)
Mà \(\widehat{ABM}=\widehat{MBN}\)
=> \(\widehat{BMN}=\widehat{MBN}\) hay \(\widehat{xBC}=\widehat{BMN}\)
b) \(Có:\widehat{BMN}=\widehat{MNx}\) (cặp góc seletrong do Bx//Ny)
Mà: \(\widehat{xBC}=\widehat{BMN}\)
=>\(\widehat{MNx}=\widehat{xBC}\) (1)
Lại có \(\widehat{xNC}=\widehat{xBC}\) (cặp góc đồng vị do Bx//By)
=>\(\widehat{MNx}=\widehat{xNC}\)
=> Nx là tia phân giác của \(\widehat{MNC}\)