Qua B kẻ BH //AK
=> \(\widehat{A}+\widehat{ABH}=180^o\)
Lại có \(\widehat{A}+\widehat{B}+\widehat{C}=360^o\)
=> \(\widehat{A}\)+\(\widehat{ABH}\)+\(\widehat{HBC}\)+\(\widehat{C}\)=360o
=> 180o+\(\widehat{HBC}\)+\(\widehat{C}\)=360o
\(\widehat{HBC}\)+\(\widehat{C}\)=180o
mà 2 góc này ở vị trí TCP
=> BH//CE
mà AK//BH( hình vẽ)
=> AK//CE(đpcm)