Giải thích các bước giải:

a.Xét $\Delta ABD,\Delta ACF$ có:

Chung $\hat A$

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

$\to \Delta ABD\sim\Delta ACF(g.g)$

b.Từ  a $\to \dfrac{AB}{AC}=\dfrac{AD}{AF}$

Mà $\widehat{DAF}=\widehat{BAC}$

$\to \Delta ADF\sim\Delta ABC(c.g.c)$

c.Xét $\Delta BHE,\Delta BDC$ có:

Chung $\hat B$

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

$\to \Delta BEH\sim\Delta BDC(g.g)$

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

$\to BH\cdot BD=BE\cdot BC$

Tương tự $CH\cdot CF=CE\cdot CB$

$\to BH\cdot BD+CH\cdot CF=BE\cdot BC+CE\cdot BC=BC^2$

Ta có:

$\dfrac{HE}{AE}+\dfrac{HD}{BD}+\dfrac{HF}{CF}=\dfrac{S_{HBC}}{S_{ABC}}+\dfrac{S_{HAC}}{S_{ABC}}+\dfrac{S_{HAB}}{S_{ABC}}=1$