Giải thích các bước giải:
Sửa đề chứng minh: $\widehat{EIF}=\dfrac12(\hat A+\hat C)$
Ta có:
$\widehat{EIF}=360^o-(\widehat{EIC}+\widehat{FIC})$
$\to \widehat{EIF}=360^o-((180^o-\widehat{IEC}-\widehat{ICE})+(180^o-\widehat{IFC}-\widehat{ICF}))$
$\to \widehat{EIF}=360^o-(360^o-\dfrac12\widehat{AED}-\dfrac12\widehat{AFB}-(\widehat{ICE}+\widehat{ICF}))$
$\to \widehat{EIF}=360^o-(360^o-\dfrac12\widehat{AED}-\dfrac12\widehat{AFB}-\widehat{BCD})$
$\to \widehat{EIF}=\dfrac12\widehat{AED}+\dfrac12\widehat{AFB}+\widehat{BCD}$
$\to \widehat{EIF}=\dfrac12(\widehat{AED}+\widehat{BCD})+\dfrac12(\widehat{AFB}+\widehat{BCD})$
$\to \widehat{EIF}=\dfrac12(\widehat{BEC}+\widehat{BCE})+\dfrac12(\widehat{DFC}+\widehat{DCF})$
$\to \widehat{EIF}=\dfrac12(180^o-\widehat{EBC})+\dfrac12(180^o-\widehat{FDC})$
$\to\widehat{EIF}=\dfrac12(360^o-\hat B-\hat D)$
$\to\widehat{EIF}=\dfrac12(\hat A+\hat C)$
