Ta có \(\widehat{A_1}=\widehat{B_3}\) mà 2 góc này ở vị trí đồng vị nên a//b
Ta có \(3\widehat{A_1}=2\widehat{A_2}\Rightarrow\dfrac{\widehat{A_1}}{2}=\dfrac{\widehat{A_2}}{3}\)
Mà \(\widehat{A_1}+\widehat{A_2}=180^0\left(kề.bù\right)\)
Áp dụng t/c dtsbn:
\(\dfrac{\widehat{A_1}}{2}=\dfrac{\widehat{A_2}}{3}=\dfrac{\widehat{A_1}+\widehat{A_2}}{2+3}=\dfrac{180^0}{5}=36^0\\ \Rightarrow\left\{{}\begin{matrix}\widehat{A_1}=72^0\\\widehat{A_2}=108^0\end{matrix}\right.\)
Ta có \(\left\{{}\begin{matrix}\widehat{A_1}=\widehat{A_3}=72^0\\\widehat{A_2}=\widehat{A_4}=108^0\end{matrix}\right.\left(đối.đỉnh\right)\)
Ta có a//b nên \(\left\{{}\begin{matrix}\widehat{A_1}=\widehat{B_1}=72^0\\\widehat{A_2}=\widehat{B_2}=108^0\end{matrix}\right.\left(so.le.trong\right);\left\{{}\begin{matrix}\widehat{A_1}=\widehat{B_3}=72^0\\\widehat{A_2}=\widehat{B_4}=108^0\end{matrix}\right.\left(đồng.vị\right)\)
