Giải thích các bước giải:

a.Xét $\Delta HAB, \Delta HAC$ có: 

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

$\widehat{BAH}=90^o-\widehat{HAC}=\widehat{HCA}$

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

b.Xét $\Delta AHB , \Delta ABC$ có:
Chung $\hat B$

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

$\to \Delta BAH\sim\Delta BCA(g.g)$

$\to \dfrac{BA}{BC}=\dfrac{BH}{BA}$

$\to BA^2=BH.BC$

Mà $BA=BE(=6)$

$\to BE^2=BH.BC$

c.Ta có $S_{ABC}=\dfrac12AB\cdot AC=24$

Vì $BD$ là phân giác $\hat B$

$\to \dfrac{DA}{DC}=\dfrac{BA}{BC}=\dfrac35$

$\to \dfrac{DA}{DA+DC}=\dfrac3{3+5}$

$\to \dfrac{DA}{AC}=\dfrac38$

$\to AD=\dfrac38AC$

$\to AD=3$

$\to S_{ABD}=\dfrac12AB\cdot AC=9$

Xét $\Delta ABD, \Delta DBE$ có:

chung $BD$

$\widehat{ABD}=\widehat{EBD}$

$BA=BE$

$\to \Delta ABD=\Delta EBD(c.g.c)$

$\to S_{ABD}=S_{EBD}$

$\to S_{DEC}=S_{ABC}-2S_{ABD}=6$