Giải thích các bước giải:
Vì $I$ là tâm đường tròn nội tiếp $\Delta ABC$
$\to AI, BI,CI$ là phân giác $\Delta ABC$
$\to \widehat{BIC}=180^o-(\widehat{IBC}+\widehat{ICB})$
$\to \widehat{BIC}=180^o-(\dfrac12\widehat{ABC}+\dfrac12\widehat{ACB})$
$\to \widehat{BIC}=180^o-\dfrac12(\widehat{ABC}+\widehat{ACB})$
$\to \widehat{BIC}=180^o-\dfrac12(180^o-\widehat{BAC})$
$\to \widehat{BIC}=180^o-(90^o-\dfrac12\widehat{BAC})$
$\to \widehat{BIC}=180^o-90^o+\dfrac12\widehat{BAC}$
$\to \widehat{BIC}=90^o+\dfrac12\widehat{BAC}$
$\to \widehat{BIC}=90^o+\dfrac12\widehat{A}$
