Giải thích các bước giải:
a.Xét $\Delta ABD,\Delta ACD$ có:
Chung $AD$
$\widehat{BAD}=\widehat{CAD}$ vì $AD$ là phân giác góc $A$
$AB=AC$
$\to\Delta ABD=\Delta ACD(c.g.c)$
$\to\widehat{ADB}=\widehat{ADC}$
Mà $\widehat{ADB}+\widehat{ADC}=\widehat{BDC}=180^o\to \widehat{ADB}=\widehat{ADC}=90^o$
$\to AD\perp BC$
b.Xét $\Delta AFI,\Delta AEI$ có:
Chung $AI$
$\widehat{FAI}=\widehat{BAD}=\widehat{CAD}=\widehat{EAI}$
$AF=AE$
$\to\Delta AFI=\Delta AEI(c.g.c)$
$\to\widehat{AFI}=\widehat{AEI}=90^o$ vì $BE\perp AC$
$\to IF\perp AB$