Giả sử: $\widehat{A_1}=80^o$
Vì $m//n$
$⇒\widehat{A_1}=\widehat{B_1}=80^o$ (đồng vị)
$⇒\widehat{A_1}=\widehat{B_4}=80^o$ (so le ngoài)
$⇒\widehat{B_4}=\widehat{A_4}=80^o$ (đồng vị)
Vì $m//n$
$⇒\widehat{A_2}+\widehat{B_1}=180^o$ (trong cùng phía bù nhau)
mà $\widehat{B_1}=80^o$
$⇒\widehat{A_2}=180^o-80^o=100^o$
$⇒\widehat{A_2}=\widehat{B_2}=100^o$ (đồng vị)
$⇒\widehat{B_2}=\widehat{A_3}=100^o$ (so le ngoài)
$⇒\widehat{A_3}=\widehat{B_3}=100^o$ (đồng vị)
Vậy $\widehat{A_3}=\widehat{A_2}=\widehat{B_3}=\widehat{B_2}=100^o$
$\widehat{A_1}=\widehat{A_4}=\widehat{B_1}=\widehat{B_4}=80^o$
