Giải thích các bước giải:
a.Vì $BE$ là phân giác $\widehat{ABC}\to\widehat{ABE}=\widehat{EBC}\to \widehat{ABE}=\widehat{EBD}$
Xét $\Delta ABE,\Delta DBE$ có:
Chung $BE$
$\widehat{ABE}=\widehat{EBD}$
$BA=BD$
$\to\Delta ABE=\Delta DBE(c.g.c)$
b.Ta có: $CH\perp BH\to BH\perp CF$
$\to \widehat{BHC}=\widehat{BHF}=90^o$
Xét $\Delta BHC,\Delta BHF$ có:
$ \widehat{BHC}=\widehat{BHF}=90^o$
Chung $BH$
$\widehat{FBH}=\widehat{ABE}=\widehat{EBC}=\widehat{HBC}$
$\to \Delta BHC=\Delta BHF(g.c.g)$
c.Ta có: $BH\perp CF, CA\perp AB\to CA\perp BF$
$BH\cap CA=E$
$\to E$ là trực tâm $\Delta FBC\to FE\perp BC$
Lại có: $\widehat{EDB}=\widehat{EAB}=90^o$ (câu a)
$\to ED\perp BC$
$\to F,E,D$ thẳng hàng