a) Có : BO,CO là p/g \(\widehat{B},\widehat{C}\)
\(\Rightarrow\)O là tâm đường tròn nội tiếp ΔABC.
\(\Rightarrow\)OA là tia p/g \(\widehat{A}\)
b) Xét tam giác ABC, có:
\(\widehat{BOC}=180^o-\widehat{OBC}-\widehat{OCB}\)
\(=180^o-\dfrac{1}{2}\widehat{ABC}-\dfrac{1}{2}\widehat{ACB}\)
\(=180^o-\dfrac{1}{2}\left(180^o-\widehat{BAC}\right)\)
\(=180^o-\dfrac{1}{2}\left(180^o-7^o\right)\)\(=93,5^o\)