Giải thích các bước giải:
a.Xét $\Delta ABF,\Delta AEC$ có:
$AE=AB$
$\widehat{EAC}=\widehat{EAB}+\widehat{BAC}=90^o+\widehat{BAC}=\widehat{FAC}+\widehat{BAC}=\widehat{FBA}$
$AB=AF$
$\to \Delta ABF=\Delta AEC(c.g.c)$
$\to BF=CE$
b.Xét $\Delta IAC, \Delta IDB$ có:
$IA=ID$
$\widehat{AIC}=\widehat{BID}$
$IC=ID$ vì $I$ là trung điểm $BC$
$\to \Delta IAC=\Delta IDB(c.g.c)$
$\to BD=AC=AF,\widehat{IAC}=\widehat{IDB}\to AC//BD$
$\to \widehat{ABD}=180^o-\widehat{BAC}=360^o-90^o-90^o-\widehat{BAC}=360^o-\widehat{EAB}-\widehat{FAC}-\widehat{BAC}=\widehat{EAF}$
Xét $\Delta AEF,\Delta BAD$ có:
$AE=AB$
$\widehat{EAF}=\widehat{ABD}$
$AF=BD$
$\to \Delta ABD=\Delta EAF(c.g.c)$
c.Gọi $AI\cap EF=H$
Từ câu b
$\to \widehat{AEF}=\widehat{BAD}$
$\to \widehat{AEH}=\widehat{BAI}$
$\to \widehat{AEH}+\widehat{EAH}=\widehat{BAI}+\widehat{EAH}=180^o-\widehat{EAB}=90^o$
$\to \Delta AEH$ vuông tại $H$
$\to AH\perp EF$
$\to AI\perp EF$