Qua $B$ kẻ đường thẳng $Bx//AM$
$Bx//AM$ mà $AM//CN$
$→Bx//CN$
$Bx//AM→\widehat A=\widehat{ABx}$ (so le trong)
$Bx//CN→\widehat C=\widehat{CBx}$ (so le trong)
$→\widehat{ABx}+\widehat{CBx}=\widehat A+\widehat C$
$→\widehat{ABC}=\widehat A+\widehat C$