a)
\(\widehat{BIC}=180^0-\widehat{ICB}-\widehat{IBC}\\ =180^0-\dfrac{1}{2}\left(\widehat{B}+\widehat{C}\right)\\ =180^0-\dfrac{1}{2}\left(180^0-\widehat{A}\right)\\ =180^0-\left(90^0-40^0\right)=180^0-50^0=130^0\)
b)\(\widehat{BIC}>\widehat{BMC}>\widehat{BAC}\) ( Áp dụng góc ngoài tam giác )