Giải thích các bước giải:

Bài 4A:

Ta có:
$\widehat{EIF}$

$=180^o-(\widehat{IFE}+\widehat{IEF})$

$=180^o-(\widehat{IFA}+\widehat{AFE}+\widehat{AEF}+\widehat{IAE})$

$=(180^o-\widehat{AFE}-\widehat{AEF})-(\widehat{IFA}+\widehat{IEA})$

$=\widehat{FAE}-\dfrac12(\widehat{AFD}+\widehat{AEB})$

$=\dfrac12(\widehat{FAE}-\widehat{AFD})+\dfrac12(\widehat{FAD}-\widehat{AEB})$

$=\dfrac12\widehat{FDA}+\dfrac12\widehat{ABE}$

$=\dfrac12(180^o-\hat D)+\dfrac12(180^o-\hat B)$

$=\dfrac12(360^o-\hat D-\hat B)$

$=\dfrac{\hat A+\hat C}2$

Bài 4B:

Kẻ $CH\perp AB, CK\perp AD$

Xét $\Delta CHB,\Delta CKD$ có:

$\hat H=\hat K(=90^o)$

$CB=CD$

$\hat B=180^o-\hat D=\widehat{CDK}$

$\to \Delta CHB=\Delta CDK$(cạnh huyền-góc nhọn)

$\to CH=CK$

Ta có: $CH\perp AB, CK\perp AD$

$\to AC$ là phân giác $\widehat{BAD}$