Xét tứ giác ABCD có :
\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)
\(\Rightarrow\widehat{B}+\widehat{C}=360-\widehat{A}-\widehat{C}\)(1)
Ta có : \(\widehat{A_1}=180-\widehat{A}\)(2 góc kề bù )
\(\widehat{C_1}=180-\widehat{C}\)(2 góc kề bù )
\(\Rightarrow\widehat{A_1}+\widehat{C_1}=180-\widehat{A}+180-\widehat{C}=360-\widehat{A}-\widehat{C}\)(2)
Từ 1 và 2 suy ra :
\(\widehat{B}+\widehat{C}=\widehat{A_1}+\widehat{C_1}\)(đpcm)