Giả sử: $\widehat{A_1}=80^o$
$m//n$
$⇒\widehat{A_1}=\widehat{B_1}=80^o$ (đồng vị)
$⇒\widehat{B_1}=\widehat{A_3}=80^o$ (so le trong)
$m//n$
$⇒\widehat{A_3}+\widehat{B_2}=180^o$ (trong cùng phía)
mà $\widehat{A_3}=80^o$
$⇒\widehat{B_2}=180^o-80^o=100^o$
$⇒\widehat{B_2}=\widehat{A_4}=100^o$ (so le trong)
$⇒\widehat{B_2}=\widehat{A_2}=100^o$ (đồng vị)
$⇒\widehat{A_4}=\widehat{B_4}=100^o$ (đồng vị)