Giải thích các bước giải:
Bài 2:
a. Do \(On \perp Oy\) nên \(\widehat{nOy}=90°\)
Ta có: \(\widehat{xOn}+\widehat{nOy}=\widehat{xOy}\)
\(\Leftrightarrow \widehat{xOn}=\widehat{xOy}-90°\)
Do \(Om \perp Ox\) nên \(\widehat{xOm}=90°\)
Ta có: \(\widehat{xOm}+\widehat{mOy}=\widehat{xOy}\)
\(\Leftrightarrow \widehat{mOy}=\widehat{xOy}-90°\)
\(\Rightarrow \widehat{mOy}=\widehat{xOn}=\widehat{xOy}-90°\)
b.
Ta có: \(\widehat{xOn}+\widehat{mOy}+\widehat{mOn}=\widehat{xOy}\)
\(\Leftrightarrow 2(\widehat{xOy}-90°)+\widehat{mOn}=\widehat{xOy}\)
\(\Leftrightarrow \widehat{mOn}=180°-\widehat{xOy}\)
\(\Leftrightarrow \widehat{xOy}+\widehat{mOn}=\widehat{xOy}+180°-\widehat{xOy}=180°\)
c.Do \(Ot\) là tia phân giác \(\widehat{xOy}\) nên:
\(\widehat{xOt}=\widehat{yOt}\)
\(\Leftrightarrow \widehat{xOn}+\widehat{nOt}=\widehat{yOm}+\widehat{mOt}\)
Mà \(\widehat{xOn}=\widehat{yOm}\) (chứng minh trên)
\(\Rightarrow \widehat{nOt}=\widehat{mOt}\)
Vậy \(Ot\) là tia phân giác \(\widehat{mOn}\)