Kẻ \(CF//AB\) mà \(AB//ED\left( {gt} \right) \Rightarrow CF//ED\)
Vì \(AB//CF \Rightarrow \widehat {BCF} = \widehat {ABC} = 30^\circ \) (so le trong)
Suy ra \(\widehat {DCF} = \widehat {DCB} - \widehat {BCF} \)\(= 100^\circ - 30^\circ = 70^\circ \)
Lại có \(ED//CF \Rightarrow \widehat {CDE} = \widehat {DCF} = 70^\circ \) (so le trong)
Vậy \(x = 70^\circ .\)
