Ta có:
$\widehat{BAC}$ + $\widehat{ACE}$ + $\widehat{CEF}$ = $360^{O}$
⇒($\widehat{BAE}$ + $\widehat{CAE}$) + $\widehat{ACE}$ + ($\widehat{CEA}$ + $\widehat{AEF}$) = $360^{O}$
⇒$\widehat{BAE}$ + $\widehat{CAE}$ + $\widehat{ACE}$ + $\widehat{CEA}$ + $\widehat{AEF}$ = $360^{O}$
⇒$\widehat{BAE}$ + ($\widehat{CAE}$ + $\widehat{ACE}$ + $\widehat{CEA}$) + $\widehat{AEF}$ = $360^{O}$
⇒$\widehat{BAE}$ + $180^{O}$ + $\widehat{AEF}$ = $360^{O}$ (tổng 3 góc tam giác)
⇒$\widehat{BAE}$ + $\widehat{AEF}$ = $180^{O}$
Mà: $\widehat{BAE}$ trong cùng phía $\widehat{AEF}$
⇒AB//EF
