Hình:
Giải:
Ta có:
\(\widehat{A_1}+\widehat{A_4}=180^0\) (Hai góc kề bù)
\(\Leftrightarrow\widehat{A_1}+2\widehat{A_1}=180^0\)
\(\Leftrightarrow3\widehat{A_1}=180^0\)
\(\Leftrightarrow\widehat{A_1}=60^0\)
\(\Leftrightarrow\widehat{A_4}=180^0-60^0=120^0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\widehat{A_1}=\widehat{A_3}=60^0\\\widehat{A_2}=\widehat{A_4}=120^0\end{matrix}\right.\) (Các cặp góc đối đỉnh)
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