Xét ΔABC có
\(\widehat{A}+\widehat{ABC}+\widehat{ACB}=180^0\)(định lí tổng ba góc trong một tam giác)
hay \(\widehat{ABC}+\widehat{ACB}=180^0-\widehat{A}=180^0-80^0=100^0\)
Ta có: \(\widehat{ABC}+\widehat{ACB}=100^0\)
hay \(2\cdot\widehat{IBC}+2\cdot\widehat{ICB}=100^0\)
\(\Leftrightarrow2\cdot\left(\widehat{IBC}+\widehat{ICB}\right)=100^0\)
\(\Leftrightarrow\widehat{IBC}+\widehat{ICB}=50^0\)
Xét ΔIBC có \(\widehat{BIC}+\widehat{IBC}+\widehat{ICB}=180^0\)(định lí tổng ba góc trong một tam giác)
hay \(\widehat{BIC}=180^0-\left(\widehat{IBC}+\widehat{ICB}\right)=180^0-50^0=100^0\)
Vậy: \(\widehat{BIC}=100^0\)
