Ta có: $\widehat{aOb}+\widehat{bOc}=180^o$ $(2$ góc kề bù) $⇒\widehat{aOm}+\widehat{mOb}+\widehat{bOn}+\widehat{nOc}=180^o$ $⇒(\widehat{aOm}+\widehat{nOc})+(\widehat{mOb}+\widehat{bOn})=180^o(1)$ Ta lại có: $\widehat{aOm}=\widehat{mOb}$ $(Om$ là phân giác $\widehat{aOb})$ $\widehat{bOn}=\widehat{nOc}$ $(On$ là phân giác $\widehat{bOc})$ $⇒\widehat{aOm}+\widehat{nOc}=\widehat{mOb}+\widehat{bOn}$ $⇒(1)⇔2.(\widehat{mOb}+\widehat{bOn})=180^o$ $⇒\widehat{mOb}+\widehat{bOn}=180^o:2=90^o$ $⇒\widehat{mOn}=90^o$
