$\widehat{AOM}=\widehat{BON}$
$→\widehat{AOM}+\widehat{BON}=2\widehat{AOM}$
$OC$ là đường phân giác $\widehat{MON}$
$→\widehat{MOC}+\widehat{NOC}=2\widehat{MOC}$
Ta có:
$\widehat{AOM}+\widehat{BON}+\widehat{MOC}+\widehat{NOC}=180^o$
$→2.\widehat{AOM}+2.\widehat{MOC}=180^o$
$→2.(\widehat{AOM}+\widehat{MOC})=180^o$
$→\widehat{AOM}+\widehat{MOC}=90^o$
mà $OM$ nằm giữa $OA,OC$
$→\widehat{AOC}=90^o$
$→OC⊥AB$
