Giải thích các bước giải:

a.Xét $\Delta BHD,\Delta BHE$ có:

Chung $BH$

$\widehat{BHD}=\widehat{BHE}(=90^o)$

$HD=HE$

$\to \Delta BHD=\Delta BHE(c.g.c)$

$\to\widehat{BEH}=\widehat{BDH}$

 $\to \widehat{BEC}=\widehat{BEH}=\widehat{BDH}=\widehat{CDA}$

b.Ta có: $AB\perp AC\to BA\perp CF$

               $BH\perp CD\to CH\perp BF$

               $BA\cap CH=D$

$\to D$ là trực tâm $\Delta FBC$

$\to FD\perp BC$

c.Xét $\Delta CHB,\Delta CHF$ có:

$\widehat{CHB}=\widehat{CHF}(=90^o)$

Chung $CH$

$\widehat{BCH}=\widehat{FCH}$ vì $CH$ là phân giác $\widehat{ACB}$

$\to \Delta CBH=\Delta CFH(g.c.g)$

$\to HB=HF$

Xét $\Delta HBE,\Delta HDF$ có:

$HE=HD$

$\widehat{EHB}=\widehat{FHD}$

$HB=HF$

$\to \Delta HEB=\Delta HDF(c.g.c)$

$\to \widehat{HEB}=\widehat{HDF}$

$\to BE//DF$