Kẻ O kẻ Oz $(\in \widehat{AOB})$ sao cho $Oz//xx'$
$→\widehat{AOz}=\widehat{OAx'}=30^{\circ}$ ( so le trong )
mà $\widehat{AOz}+\widehat{zOB}=100^{\circ}$
$→\widehat{zOB}=100^{\circ}-30^{\circ}=70^{\circ}$
$→\widehat{zOB}+\widehat{OBy}=70^{\circ}+110^{\circ} = 180{\circ}$
$→Oz//yy'$
$\to \left\{\begin{matrix}
xx'//Oz & \\
yy'//Oz &
\end{matrix}\right.$
$→xx'//yy'$
