Giải thích các bước giải:
a.Ta có $\widehat{AOB}$ tù $,\widehat{AOC}=\widehat{DOB}=90^o$
$\to \widehat{AOC}=\widehat{DOB}<\widehat{AOB}$
$\to OD,OC$ nằm giữa $OA,OB$
$\to \widehat{AOD}=\widehat{AOB}-\widehat{BOD}=\widehat{AOB}-90^o=\widehat{AOB}-\widehat{AOC}=\widehat{BOC}$
b.Vì $OM,ON$ là phân giác $\widehat{BOC},\widehat{AOD},\widehat{AOD}=\widehat{BOC}$
$\to \widehat{MOC}=\dfrac12\widehat{BOC}=\dfrac12\widehat{AOD}=\widehat{DON}$
$\to \widehat{MON}=\widehat{MOC}+\widehat{COD}+\widehat{DON}$
$\to \widehat{MON}=\widehat{MOC}+\widehat{DON}+\widehat{COD}$
$\to \widehat{MON}=\widehat{DON}+\widehat{DON}+\widehat{COD}$
$\to \widehat{MON}=2\widehat{DON}+\widehat{COD}$
$\to \widehat{MON}=\widehat{DOA}+\widehat{COD}$
$\to \widehat{MON}=\widehat{AOC}$
$\to \widehat{MON}=90^o$
$\to OM\perp ON$
