Bài 3
Ta có a// b
=> \(\widehat{A_1}+\widehat{B_1}=180^o\)
mà \(\widehat{A_1}=\widehat{B_1}=>\widehat{A_1}=\widehat{B_1}=90^o\)
hay \(c\perp a;c\perp b\)
Bài 4
ta có AB // CD
\(=>\widehat{A}=\widehat{ACD}=\left(\widehat{BCD}-\widehat{ACB}\right)=115-40=75^o\\ \)
áp dụng tổng 3 góc trong tam giác
\(\widehat{A}+\widehat{B}+\widehat{ACB}=180^o=>\widehat{B}=180-40-75=65^o\)