Giải thích các bước giải:
Bài 6:
Ta có $Om, On$ là phân giác $\widehat{xOy},\widehat{yOz}$
$\to\widehat{mOn}=\widehat{mOy}+\widehat{yOn}=\dfrac12\widehat{xOy}+\dfrac12\widehat{yOz}=\dfrac12(\widehat{xOy}+\widehat{yOz})=\dfrac12\cdot 180^o=90^o$
$\to Om\perp On$
Bài 7:
Ta có: $Om\perp On$
$\to \widehat{yOn}=90^o-\widehat{yOm}=90^o-\dfrac12\widehat{xOy}=\dfrac12(180^o-\widehat{xOy})=\dfrac12\widehat{yOz}$
Lại có:
$\widehat{zOn}=180^o-\widehat{xOm}-\widehat{mOn}=180^o-\dfrac12\widehat{xOy}-90^o=90^o-\dfrac12\widehat{xOy}=\dfrac12(180^o-\widehat{xOy})=\dfrac12\widehat{yOz}$
$\to \widehat{yOn}=\widehat{nOz}=\dfrac12\widehat{yOz}$
$\to On$ là phân giác $\widehat{yOz}$
