Ta có \(\widehat{A_4}=\widehat{A_1}=60^0\left(đối.đỉnh\right)\)
\(\widehat{B_1}+\widehat{B_5}=180^0\left(kề.bù\right)\\ \Rightarrow\widehat{B_5}=180^0-100^0=80^0\)
Vì a//b nên \(\widehat{C_1}=\widehat{A_4}=60^0\left(đồng.vị\right);\left\{{}\begin{matrix}\widehat{B_1}=\widehat{D_2}=100^0\\\widehat{B_5}=\widehat{D_3}=80^0\end{matrix}\right.\left(đồng.vị\right)\)