Giải:
tam giác ABC có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)
=> \(\widehat{B}+\widehat{C}=180^o-\widehat{A}=180^o-72^o=108^o\)
Tam giác IBC có: \(\widehat{BIC}+\widehat{B_1}+\widehat{C_1}=180^o\)
=> \(\widehat{BIC}=180^o-\left(\widehat{B_1}+\widehat{C_1}\right)\)
= \(180^o-\left(\dfrac{1}{2}\widehat{B}+\dfrac{1}{2}\widehat{C}\right)=180^o-\dfrac{1}{2}\left(\widehat{B}+\widehat{C}\right)\)
\(=180^o-\dfrac{1}{2}\cdot108^o=180^o-54^o=126^o\)
Xét tam giác ABC có: \(\widehat{A}+\widehat{ABC}+\widehat{ACB}=180^0\) (Đl tổng 3 góc trong tam giác)
\(\Rightarrow\widehat{ABC}+\widehat{ACB}=180^0-\widehat{A}=180^0-72^0=108^0\)
Vì I là giao điểm của 2 tia phân giác của góc ABC và góc ACB nên:\(\widehat{IBC}+\widehat{ICB}=\dfrac{\widehat{ABC}}{2}+\dfrac{\widehat{ACB}}{2}=\dfrac{\widehat{ABC}+\widehat{ACB}}{2}=\dfrac{108^0}{2}=54^0\)
Xét tam giác IBC có:
\(\widehat{BIC}+\widehat{IBC}+\widehat{ICB}=180^0\)(ĐL tổng 3 góc trong tam giác)
\(\Rightarrow\widehat{BIC}=180^0-\left(\widehat{IBC}+\widehat{ICB}\right)=180^0-54^0=126^0\)
