$\widehat{xOM}+\widehat{MON}+\widehat{NOy}=180^o$
$\Rightarrow \widehat{MON}=180^o-\widehat{MOx}-\widehat{NOy}$
$\widehat{EOx}+\widehat{EOF}+\widehat{FOy}=180^o$
$\Rightarrow \widehat{EOF}=180^o-\widehat{EOx}-\widehat{FOy}$
Mà $\widehat{MOx}=\widehat{EOx}$, $\widehat{NOy}=\widehat{FOy}$
$\Rightarrow \widehat{MON}=\widehat{EOF}$
$OE\bot OF\Rightarrow \widehat{EOF}=90^o$
Vậy $\widehat{MON}=90^o$