Tạm 2 cách
Cách 1:
Qua \(C\) kẻ \(Cz//Ax\)
\(Cz//Ax\) mà \(Ax//By\)
\(→Cz//By\)
\(Cz//Ax→\widehat{A}+\widehat{ACz}=180^\circ\) (hai góc trong cùng phía)
mà \(\widehat A=60^\circ\)
\(→\widehat{ACz}=180^\circ-60^\circ=120^\circ\)
mà \(\widehat{BCz}+\widehat{ACB}=\widehat{ACz}\)
\(→\widehat{BCz}+40^\circ=120^\circ\)
\(↔\widehat{BCz}=80^\circ\)
\(Cz//By→\widehat{BCz}+\widehat{CBy}=180^\circ\) (hai góc trong cùng phía)
mà \(\widehat{BCz}=80^\circ\)
\(→\widehat{CBy}=180^\circ-80^\circ=100^\circ\)
Vậy \(\widehat{CBy}=100^\circ\)
Cách 2:
Qua \(B\) kẻ đường thẳng \(Dz//AC(D∈Ax)\)
\(Dz//AC\)
\(→\widehat{ACB}=\widehat{CBz}=40^\circ\) (hai góc so le trong)
và \(\widehat{CAx}=\widehat{BDx}=60^\circ\) (hai góc đồng vị)
\(Ax//By→\widehat{zBy}=\widehat{BDx}=60^\circ\) (hai góc đồng vị)
Ta có: \(\widehat{zBy}+\widehat{CBz}=\widehat{CBy}\)
\(→60^\circ+40^\circ=\widehat{CBy}\) hay \(100^\circ=\widehat{CBy}\)
Vậy \(\widehat{CBy}=100^\circ\)
