Giải thích các bước giải:

a.Xét $\Delta ACF,\Delta ABE$ có:

Chung $\hat A$

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

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

b.Xét $\Delta HBF,\Delta HCE$ có:

$\widehat{FHB}=\widehat{EHC}$ (đối đỉnh)

$\widehat{HFB}=\widehat{HEC}(=90^o)$

$\to \Delta HBF\sim\Delta HCE(g.g)$

$\to \dfrac{HB}{HC}=\dfrac{HF}{HE}$

$\to HE\cdot HB=HF\cdot HC$

c.Xét $\Delta AEF,\Delta ABC$ có:

Chung $\hat A$

$\dfrac{AE}{AB}=\dfrac{AF}{AC}$ vì $AE\cdot AC=AF\cdot AB$

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

$\to \widehat{AEF}=\widehat{ABC}$

d.Gọi $AH\cap BC=D$

Vì $BE\perp AC, CF\perp AB, BE\cap CF=H$

$\to H$ là trực tâm $\Delta ABC$

$\to AH\perp BC\to AD\perp BC$

Xét $\Delta BDH,\Delta BCE$ có:

Chung $\hat B$

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

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

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

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

Tương tự chứng minh được $CH\cdot CF=CD\cdot CB$

$\to BH\cdot BE+CH\cdot CF=BD\cdot BC+CD\cdot CB=(BD+DC)\cdot BC=BC\cdot BC=BC^2$