Question

cho tam giác ABC , hai tia phân giác của  góc b và góc c cắt tại I . cmr góc BIC  = 90 độ + góc A/2

Answer 1

Xét ΔABC có 

\(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)

\(\Leftrightarrow2\cdot\left(\widehat{IBC}+\widehat{ICB}\right)=180^0-\widehat{A}\)

\(\Leftrightarrow\widehat{IBC}+\widehat{ICB}=\dfrac{180^0-\widehat{A}}{2}\)

Xét ΔIBC có 

\(\widehat{BIC}+\widehat{IBC}+\widehat{ICB}=180^0\)

\(\Leftrightarrow\widehat{BIC}=180^0-90^0+\dfrac{\widehat{A}}{2}\)

\(\Leftrightarrow\widehat{BIC}=90^0+\dfrac{\widehat{A}}{2}\)