Giải:
Kẻ EF // AB ( \(EF\in\widehat{AEB}\))
\(\widehat{A}=\widehat{E_1=}45^o\) ( so le trong )
Vì AC _|_CD, BD _|_ CD nên AC // BD
Vì BC // AC, AC // EF nên BC // EF
\(\Rightarrow\widehat{B}=\widehat{E_2}=60^o\)
\(\Rightarrow\widehat{AEB}=\widehat{E_1}+\widehat{E_2}\)
\(\Rightarrow\widehat{AEB}=45^o+60^o=105^o\)
Vậy \(\widehat{AEB}=105^o\)