Có \(\widehat{A_1}+\widehat{A_2}+\widehat{B_1}=288^0\)
\(\Leftrightarrow180^0+\widehat{B_1}=288^0\)
\(\Leftrightarrow\widehat{B_1}=108^0\)
\(\Rightarrow\widehat{B_2}=180^0-108^0=72^0\)
Có \(\left\{{}\begin{matrix}\widehat{A_1}+\widehat{A_2}=180^0\\\widehat{A_1}=\dfrac{2}{3}\widehat{A_2}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}\widehat{A_2}+\widehat{A_2}=180^0\\\widehat{A_1}=\dfrac{2}{3}\widehat{A_2}\end{matrix}\right.\)\(\Rightarrow\widehat{A_2}=108^0\)
\(\Rightarrow\widehat{A_3}=180^0-108^0=72^0\)
\(\Rightarrow\widehat{B_2}=\widehat{A_3}\)
mà hai góc nằm ở vị trí đồng vị nên a//b
theo hình vẽ \(=>\angle\left(A2\right)+\angle\left(A3\right)=180^o\)(1)
mà \(\angle\left(A1\right)=\dfrac{2}{3}\angle\left(A2\right)=>\angle\left(A2\right)=\dfrac{3}{2}\angle\left(A1\right)\)
\(1,5\angle\left(A1\right)+\angle\left(A3\right)=180^o\) do \(\angle\left(A1\right)=\angle\left(A3\right)\)(đối đỉnh)
\(=>2,5\angle\left(A1\right)=180=>\angle\left(A1\right)=72^0\)
\(=>\angle\left(A2\right)=\dfrac{3}{2}.72=108^o\)=>\(\angle\left(B1\right)=288-108-72=108^o\)
\(=>\angle\left(B2\right)=180-\angle\left(B1\right)=72^o\)
từ(1) =>\(\angle\left(A3\right)=180-\angle\left(A2\right)=180-108=72^o=\angle\left(B2\right)\)
do 2 góc này ở vị trí so le trong=>a//b
