a) Ta có: $\widehat{ABE}+\widehat{EBC}=180^{\circ}$ (2 góc kề bù)
=> $110+\widehat{EBC}=180^{\circ}$
=> $\widehat{EBC}=180^{\circ}-110^{\circ}=70^{\circ}$
Xét tam giác EBC: $\widehat{E}+\widehat{B}+\widehat{C}=180^{\circ}$ (Tổng ba góc trong tam giác)
=> $40^{\circ}+70^{\circ}+\widehat{C}=180^{\circ}$
=> $\widehat{C}=180^{\circ}-110^{\circ}=70^{\circ}$
=> $\widehat{EBC}=\widehat{ECB}=70^{\circ}$
=> Tam giác EBC cân tại E => EB = EC
b) Ta có: $\widehat{ECD}=180^{\circ}-\widehat{ECB}=180^{\circ}-70^{\circ}=110^{\circ}=>\widehat{ABE}=\widehat{DCE}$
Xét tam giác ABE và DCE có:
BE = CE (cmt)
$\widehat{ABE}=\widehat{DCE}$ (cmt)
AB = DC (gt)
=> $\Delta ABE=\Delta ACE$ (c.g.c) => AE = DE