Xét ΔABC có \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)
=>\(2\cdot\left(\widehat{IBC}+\widehat{ICB}\right)+60^0=180^0\)
=>\(2\cdot\left(\widehat{IBC}+\widehat{ICB}\right)=120^0\)
=>\(\widehat{IBC}+\widehat{ICB}=60^0\)
Xét ΔIBC có \(\widehat{IBC}+\widehat{ICB}+\widehat{BIC}=180^0\)
=>\(\widehat{BIC}+60^0=180^0\)
=>\(\widehat{BIC}=120^0\)