Giải thích các bước giải:

a.Xét $\Delta ABE,\Delta ACF$ có:

Chung $\hat A$

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

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

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

$\to AE\cdot AC=AF\cdot AB$

b.Xét $\Delta AEF, \Delta ABC$ có:

Chung $\hat A$

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

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

c.Xét $\Delta HBF,\Delta HCE$ có:

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

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

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

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

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

Mà $\widehat{FHE}=\widehat{BHC}$

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

d. Tương tự câu a chứng minh được $BD\cdot BC=BF\cdot BA, CE\cdot CA=CD\cdot CB$

$\to BF\cdot BA+CE\cdot CA=BD\cdot bC+CD\cdot BC=(BD+DC)\cdot BC=BC\cdot BC$