Giải thích các bước giải:

a.Ta có: $BE\perp AC, CF\perp AB, BE\cap CF=H\to H$ là trực tâm $\Delta ABC$

$\to AH\perp BC$

b.Xét $\Delta ABE,\Delta ACF$ có:

Chung $\hat A$

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

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

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

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

c.Xét $\Delta AEF,\Delta ABC$ có:

Chung $\hat A$ 

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

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

Tương tự chứng minh được $\Delta BFD\sim\Delta BCA(c.g.c)$

d.Xét $\Delta CHD,\Delta CFB$ có:

Chung $\hat C$

$\widehat{CDH}=\widehat{CFB}(=90^o)$

$\to \Delta CDH\sim\Delta CFB(g.g)$

$\to \dfrac{CD}{CF}=\dfrac{CH}{CB}$

$\to \dfrac{CD}{CH}=\dfrac{CF}{CB}$

Mà $\widehat{HCD}=\widehat{FCB}$

$\to \Delta CBH\sim\Delta CFD(c.g.c)$

e.Từ câu c $\to \widehat{AFE}=\widehat{ACB}$

Xét $\Delta IBF,\Delta ICE$ có:

Chung $\hat I$

$\widehat{IFB}=\widehat{AFE}=\widehat{ACB}=\widehat{ICE}$

$\to \Delta IBF\sim\Delta IEC(g.g)$

$\to \dfrac{IB}{IE}=\dfrac{IF}{IC}$

$\to IE\cdot IF=IB\cdot IC$

g.Từ câu c ta có: $\Delta AEF\sim\Delta ABC, \Delta BFD\sim\Delta BCA$

$\to\Delta AEF\sim\Delta DBF$

$\to \widehat{AFE}=\widehat{BFD}$

$\to  90^o-\widehat{AFE}=90^o-\widehat{BFD}$

$\to \widehat{EFC}=\widehat{CFD}$

$\to FC$ là phân giác $\widehat{EFD}$