Giải thích các bước giải:

a.Xét $\Delta HAB,\Delta ABC$ có:

Chung $\hat B$

$\widehat{AHB}=\widehat{BAC}(=90^o)$

$\to \Delta HBA\sim\Delta ABC(g.g)$

Ta có:

$\Delta ABC$ vuông tại $A$

$\to BC^2=AB^2+AC^2$

$\to BC=\sqrt{AB^2+AC^2}$

$\to BC=15$

Mà $AH\perp BC$

$\to AH.BC=AB.AC(=2S_{ABC})$

$\to AH=\dfrac{AB.AC}{BC}= 7.2$

b.Xét $\Delta AHB,\Delta AHC$ có:
$\widehat{AHB}=\widehat{AHC}(=90^o)$

$\widehat{HAB}=90^o-\widehat{HAC}=\hat C$

$\to \Delta HAB\sim\Delta HCA(g.g)$

$\to \dfrac{AH}{HC}=\dfrac{HB}{AH}$

$\to AH^2=BH.HC$

c.Xét $\Delta ADH,\Delta AHB$ có:
Chung $\hat A$

$\widehat{ADH}=\widehat{AHB}(=90^o)$

$\to \Delta AHD\sim\Delta ABH(g.g)$

$\to \dfrac{AH}{AB}=\dfrac{AD}{AH}$

$\to AH^2=AB.AD$

Tương tự: $AH^2=AC.AE$

$\to AB.AD=AC.AE$

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

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

d.Ta có:

$BH=\sqrt{AB^2-AH^2}=\sqrt{9^2-7.2^2}=5.4$

Xét $\Delta ABK,\Delta FBH$ có:

$\widehat{BAK}=\widehat{BHF}(=90^o)$

$\widehat{ABK}=\widehat{HBF}$ vì $BF$ là phân giác $\hat B$

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

$\to \dfrac{S_{ABK}}{S_{HBF}}=\dfrac{AB^2}{HB^2}=\dfrac{9^2}{5.4^2}=\dfrac{25}9$