Question

Cho tam giác ABC  có \(\widehat{A}=70^0\), các đường phân giác BD, CE cắt nhau ở I. Tính \(\widehat{BIC}\) ?

Answer 1

\(\Delta ABC\) có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\) (định lí)

\(\widehat{B}+\widehat{C}=180^o-\widehat{A}=180^o-70^o\)

\(\Rightarrow\) \(\widehat{B}+\widehat{C}=110^o\).

Do \(\widehat{B_1}=\widehat{B_2},\widehat{C_1}=\widehat{C_2}\) nên \(\widehat{B_1}+\widehat{C_1}=\dfrac{\widehat{B}+\widehat{C}}{2}=\dfrac{110^o}{2}=55^o\)

Vậy: \(\widehat{BIC}=180^o-\left(\widehat{B_1}+\widehat{C_1}\right)=180^o-55^o=125^o.\)