Giải thích các bước giải:

a.Xét $\Delta ADC,\Delta BCE$ có:

Chung $\hat C$

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

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

b.Vì $AD\perp BC, BE\perp AC, AD\cap BE=H$

$\to H$ là trực tâm $\Delta ABC$

$\to CH\perp AB=F$

$\to \widehat{AFC}=\widehat{AEB}(=90^o)$

Mà $\widehat{EAB}=\widehat{FAC}$

$\to \Delta AEB\sim\Delta AFC(g.g)$

$\to \dfrac{AE}{AF}=\dfrac{AB}{AC}$

$\to AF.AB=AE.AC$

b.Ta có:

$\dfrac{HD}{AD}=\dfrac{\dfrac12HD.BC}{\dfrac12AD.BC}=\dfrac{S_{HBC}}{S_{ABC}}$

Tương tự:
$\dfrac{HE}{BE}=\dfrac{S_{HAC}}{S_{ABC}}$

$\dfrac{HF}{CF}=\dfrac{S_{HAB}}{S_{ABC}}$

$\to \dfrac{HD}{AD}+\dfrac{HE}{BE}+\dfrac{HF}{CF}=\dfrac{S_{HBC}+S_{HAC}+S_{HAB}}{S_{ABC}}=\dfrac{S_{ABC}}{S_{ABC}}=1$