Giải thích các bước giải:

a.Xét $\Delta HBF,\Delta HCE$ có:

$\hat F=\hat E(=90^o)$

$\widehat{FHB}=\widehat{EHC}$

$\to \Delta HFB\sim\Delta HEC(g.g)$

b.Từ a $\to \dfrac{HF}{EH}=\dfrac{HB}{HC}$

$\to HE.HB=HF.HC$

Tương tự: $HD.HA=HF.HC$

$\to HA.HD=HB.HE=HC.HF$

c.Xét $\Delta HEF, \Delta HCB$ có:

$\dfrac{HE}{HC}=\dfrac{HF}{HB}$ vì $HE.HB=HF.HC$

$\widehat{EHF}=\widehat{BHC}$

$\to \Delta HEF\sim\Delta HCB(c.g.c)$

$\to \widehat{HEF}=\widehat{HCB}$

Xét $\Delta BDH,\Delta BEC$ có:

Chung $\hat B$

$\widehat{BDH}=\widehat{BEC}(=90^o)$

$\to \Delta BDH\sim\Delta BEC(g.g)$

$\to \dfrac{BD}{BE}=\dfrac{BH}{BC}$

$\to \Delta BDE\sim\Delta BHC(c.g.c)$

$\to \widehat{BED}=\widehat{BCH}=\widehat{HEF}$

$\to EH$ là phân giác $\widehat{FED}$