a.
Ta có \(\widehat{D_2}=\widehat{D_1}\) (hai góc đối đỉnh)
Mà \(\widehat{D_1}=50^0\)
Nên \(\widehat{D_2}=50^0\)
b.
Do \(\left\{{}\begin{matrix}a\perp c\\b\perp c\end{matrix}\right.\)
Suy ra \(a||b\)
c.
Do \(a||x\) nên \(\widehat{CEx}=\widehat{C_1}\) (hai góc so le trong)
Mà \(\widehat{C_1}=40^0\)
Suy ra \(\widehat{CEx}=40^0\)
Do \(\left\{{}\begin{matrix}a||x\\a||b\end{matrix}\right.\) nên \(b||x\)
Suy ra \(\widehat{DEx}=\widehat{D_1}\) (hai góc so le trong)
Suy ra \(\widehat{DEx}=50^0\)
Vậy \(\widehat{CED}=\widehat{CEx}+\widehat{DEx}=40^0+50^0=90^0\)