Giải:
Kẻ EF // AB
\(\Rightarrow\widehat{BAE}+\widehat{AEF}=180^o\) ( cặp góc trong cùng phía )
\(\Rightarrow60^o+\widehat{AEF}=180^o\)
\(\Rightarrow\widehat{AEF}=120^o\)
Vì AB // EF, AB // CD nên CD // EF
\(\Rightarrow\widehat{CEF}+\widehat{DCE}=180^o\) ( cặp góc trong cùng phía )
\(\Rightarrow\widehat{CEF}+110^o=180^o\)
\(\Rightarrow\widehat{CEF}=70^o\)
Ta có: \(\widehat{AEC}=\widehat{AEC}+\widehat{CEF}\)
\(\Rightarrow120^o=\widehat{AEC}+70^o\)
\(\Rightarrow\widehat{AEC}=50^o\)
Vậy \(\widehat{AEC}=50^o\)
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