Đáp án:
$\begin{array}{l}
Cho:\widehat {BAC} = {30^0};\widehat {ABC} = {65^0}\\
Do:\widehat {BAC} + \widehat {ABC} + \widehat {BCA} = {180^0}\\
\Leftrightarrow \widehat {BCA} = {180^0} - {30^0} - {65^0} = {85^0}\\
a//b \Leftrightarrow \left\{ \begin{array}{l}
\widehat {CDE} = \widehat {BAC} = {30^0}\left( {so\,le\,trong} \right)\\
\widehat {CED} = \widehat {ABC} = {65^0}\left( {so\,le\,trong} \right)\\
\widehat {DCE} = \widehat {BCA} = {85^0}
\end{array} \right.
\end{array}$
