a) Kẻ Bz // Ax suy ra Bz // Cy

Ta có : $\widehat{BAx}+\widehat{ABz}=180{}^\circ $ ( Hai góc trong cùng phía )

$\widehat{BCy}+\widehat{CBz}=180{}^\circ $ ( Hai góc trong cùng phía )

Suy ra $\widehat{BAx}+\widehat{ABz}+$$\widehat{BCy}+\widehat{CBz}=180{}^\circ +180{}^\circ =360{}^\circ $ (1)

Mà $\widehat{ABC}=\widehat{ABz}+\widehat{CBz}$ nên từ (1) ta được $\widehat{BAx}+\widehat{ABC}+\widehat{BCy}=360{}^\circ \Leftrightarrow \widehat{A}+\widehat{B}+\widehat{C}=360{}^\circ $

b) Kẻ Bz // Ax Ta có : $\widehat{BAx}+\widehat{ABz}=180{}^\circ $ ( Hai góc trong cùng phía )

$\begin{align}& \widehat{A}+\widehat{B}+\widehat{C}=360{}^\circ \Leftrightarrow \widehat{BAx}+\widehat{ABC}+\widehat{BCy}=360{}^\circ \\ & \\ \end{align}$

Mà $\widehat{ABC}=\widehat{ABz}+\widehat{CBz}$ nên $\widehat{BAx}+\widehat{ABz}+$$\widehat{CBz}+\widehat{BCy}=360{}^\circ $

Hay $180{}^\circ +\widehat{CBz}+\widehat{BCy}=360{}^\circ $$\Rightarrow $$\widehat{CBz}+\widehat{BCy}=360{}^\circ -180{}^\circ =180{}^\circ $

Hai góc $\widehat{CBz}$ và $\widehat{BCy}$ ở vị trí trong cùng phía nên Bz // Cy
Vậy Ax // Bz // Cy hay Ax // Cy