Đáp án: $\widehat{BIC}=90^o+\dfrac12\widehat{BAC}$
Giải thích các bước giải:
Ta có BI,CI là phân giác $\widehat{ABC},\widehat{ACB}$
$\begin{split}\to \widehat{BIC}&=180^o-(\widehat{IBC}+\widehat{ICB})\\&=180^o-(\dfrac12\widehat{ABC}+\dfrac12\widehat{ACB})\\&=180^o-\dfrac12(\widehat{ABC}+\widehat{ACB})\\&=180^o-\dfrac12(180^o-\widehat{BAC})\\&=180^o-90^o+\dfrac12\widehat{BAC}\\&=90^o+\dfrac12\widehat{BAC}\end{split}$