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