Bài giải:

Ta có: $\widehat{ABI}=\widehat{IBC}=\frac{\widehat{ABC}}{2}$

$\widehat{ACI}=\widehat{ICB}=\frac{\widehat{ACB}}{2}$

$\Rightarrow \widehat{IBC}+\widehat{ICB}=\frac{\widehat{ABC}}{2}+\frac{\widehat{ACB}}{2}=\frac{{{180}^{0}}-\widehat{A}}{2}=\frac{{{180}^{0}}-{{70}^{0}}}{2}={{55}^{0}}$

Tam giác BIC: $\widehat{BIC}+\widehat{ICB}+\widehat{IBC}={{180}^{0}}$

$\Rightarrow \widehat{BIC}={{180}^{0}}-\left( \widehat{ICB}+\widehat{IBC} \right)={{180}^{0}}-{{55}^{0}}={{125}^{0}}$

Đáp án đúng là C