Giải thích các bước giải:

a.Xét $\Delta HAB, \Delta ABC$ có:

Chung $\hat B$

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

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

b.Từ câu a

$\to \dfrac{BA}{BC}=\dfrac{HB}{AB}$

$\to BA^2=BH.BC$

Mà $BA=BE\to BE^2=BH.BC$

c.Ta có $\Delta ABC$ vuông tại $A\to BC^2=AB^2+AC^2=25\to BC=5$

Mà $AH\perp BC$

$\to S_{ABC}=\dfrac12AB.AC=\dfrac12AH.BC$

$\to AH=\dfrac{AB.AC}{BC}=\dfrac{12}{5}$

d.Vì $BD$ là phân giác $\hat B$

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

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

$\to \dfrac{AC}{CD}=\dfrac85$

$\to CD=\dfrac58AC=\dfrac52$

Xét $\Delta ABD, \Delta EBD$ có:

$BA=BE$

$\widehat{ABD}=\widehat{DBE}$

Chung $BD$

$\to\Delta BAD=\Delta BED(c.g.c)$

$\to \widehat{DEB}=\widehat{DAB}=90^o\to DE\perp BC\to \widehat{CED}=90^o$

Xét $\Delta CDE,\Delta CAB$ có:

Chung $\hat C$

$\widehat{CED}=\widehat{CAB}(=90^o)$

$\to\Delta CDE\sim\Delta CBA(g.g)$

$\to \dfrac{S_{CDE}}{S_{CBA}}=(\dfrac{CD}{CB})^2=\dfrac14$