Ta có \(Ox\) là tia đối ủa tia OA (vì OA = OB = OC)
=> ΔOAB ; ΔOAC cân tại O
\(=>\widehat{OAB}=\widehat{OBA};\widehat{OAC}=\widehat{OCA}\)
Vì \(\widehat{BOx};\widehat{COx}\) là 2 góc ngoài của ΔAOB và ΔAOC
\(=>\left\{{}\begin{matrix}\widehat{BOx}=\widehat{AOB}+\widehat{OBA}=2\widehat{OAB}\\\widehat{COx}=\widehat{OAC}+\widehat{OCA}=2\widehat{OAC}\end{matrix}\right.\)
\(=>\widehat{BOC}=\widehat{BOx}+\widehat{COx}=2\widehat{OAB}+2\widehat{OAC}=2\widehat{A}\)
Có \(\widehat{BIC}=160^o-\widehat{IBC}-\widehat{ICB}=180^o-\dfrac{\widehat{B}}{2}-\dfrac{\widehat{C}}{2}\)
\(=180^o-\left(90^o-\dfrac{\widehat{A}}{2}\right)=>4\widehat{BIC}-\widehat{BOC}=360^o+2\widehat{A}-2\widehat{A}=360^o\)
