Xét \(\Delta ABC\)có \(\widehat{A}=40^0\);\(\widehat{B}=70^0\)
Ta có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)(t/c tổng 3 góc)
\(\Rightarrow40^0+70^0+\widehat{C}=180^0\)
\(\Rightarrow110^0+\widehat{C}=180^0\)
\(\Rightarrow\widehat{C}=180^0-110^0\)
\(\Rightarrow\widehat{C}=70^0=\widehat{B}\)
Vậy bài toán được chứng minh