Giải thích các bước giải:

a.Xét $\Delta AEB,\Delta ACF$ có:

chung $\hat A$

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

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

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

c.Xét $\Delta AEF,\Delta ABC$ có:

Chung $\hat A$

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

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

d.Gọi $AH\cap BC=D$

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

$\to AD\perp BC$

Xét $\Delta BHD,\Delta BCE$ có:
Chung $\hat B$

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

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

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

$\to BH.BE=BD.BC$

Tương tự chứng minh được $CH.CF=CD.CB$

$\to BH.BE+CH.CF=BD.BC+CD.CB=BC^2$

e.Ta có $EB\perp AC\to AB=\sqrt{AE^2+EB^2}=5$

Từ câu c

$\to \dfrac{S_{AEF}}{S_{ABC}}=(\dfrac{AE}{AB})^2=\dfrac9{25}$

f.Xét $\Delta BDH, \Delta ADC$ có:

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

$\widehat{BHD}=90^o-\widehat{HBD}=90^o-\widehat{EBC}=\widehat{ECB}=\widehat{ACD}$

$\to \Delta DBH\sim\Delta DCA(g.g)$
$\to \dfrac{DB}{DC}=\dfrac{DH}{DA}$

$\to DH.DA=DB.DC\le\dfrac14(DB+DC)^2=\dfrac14BC^2$