Giải thích các bước giải:
a.Xét $\Delta ABK,\Delta ADE$ có:
$AB=AD$
$\widehat{ABK}=\widehat{ADE}=90^o$
$BK=DE$
$\to\Delta ABK=\Delta ADE(c.g.c)$
$\to\widehat{KAB}=\widehat{EAD}$
$\to\widehat{EAK}=\widehat{EAB}+\widehat{BAK}=\widehat{EAB}+\widehat{DAE}=\widehat{DAB}=90^o$
Vì $\Delta AEF$ vuông cân tại $E$
$\to\widehat{EAF}=45^o\to\widehat{EAI}=45^o$
$\to\widehat{IAK}=\widehat{EAK}-\widehat{EAI}=45^o$
b.Từ câu a
$\to\widehat{EAI}=\widehat{IAK}(=45^o), AE=AK$
Xét $\Delta AEI,\Delta AKI$ có:
Chung $AI$
$\widehat{EAI}=\widehat{IAK}$
$AE=AK$
$\to\Delta AEI=\Delta AKI(c.g.c)$
$\to\widehat{AEI}=\widehat{AKI}$
$\to\widehat{AEI}=\widehat{AKB}$
Mà $\widehat{AKB}=\widehat{AED}$ (câu a)
$\to \widehat{AEI}=\widehat{AED}$
$\to EA$ là phân giác $\widehat{IED}$
