Ta có: \(\widehat{NBK}=\widehat{NBA};\widehat{ABO}=\widehat{OBC}\)
Mà \(\widehat{NBK}+\widehat{NBA}+\widehat{ABO}+\widehat{OBC}=180^o\)
\(\Rightarrow\widehat{NBA}+\widehat{NBA}+\widehat{ABO}+\widehat{OBA}=180^o\)
\(\Rightarrow2\widehat{NBA}+2\widehat{OBA}=180^o\Leftrightarrow\widehat{NBA}+\widehat{OBA}=90^o\text{}\text{}\)
Tương tự: \(\widehat{ACM}+\widehat{OCA}=90^o\text{}\text{}\)
Ta có: \(\widehat{BNC}=90^o-\widehat{BON};\widehat{BMC}=90^o-\widehat{COM}\)
Mà \(\widehat{BON}=\widehat{COM}\) (2 góc đối đỉnh)
\(\Rightarrow\widehat{BNC}=\widehat{BMC}\)
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