Giải thích các bước giải:

a.Xét $\Delta ABC,\Delta HBA$ có:

Chung $\hat B$

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

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

b.Xét $\Delta AHD,\Delta CDE$ có:

$\widehat{ADH}=\widehat{CDE}$(đối đỉnh)

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

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

$\to \dfrac{AH}{CE}=\dfrac{AD}{CD}$

$\to AH\cdot CD=AD\cdot CE$

c.Từ câu b $\to \dfrac{DA}{DC}=\dfrac{DH}{DE}$

$\to \dfrac{DA}{DH}=\dfrac{DC}{DE}$

Mà $\widehat{HDE}=\widehat{ADC}$

$\to \Delta DHE\sim\Delta DAC(c.g.c)$

d.Ta có: $\Delta ABC$ vuông tại $A$

$\to BC=\sqrt{AB^2+AC^2}=10$

Mà $AH\perp BC$

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

$\to AH=\dfrac{AB\cdot AC}{BC}=4.8$

$\to BH=\sqrt{AB^2-AH^2}=3.6$

$\to DH=HB=3.6$

$\to DB=BH+HD=7.2$

$\to CD=BC-BD=2.8$

Ta có: $AD^2=AH^2+HD^2=AH^2+HB^2=AB^2$

$\to AD=AB=6$

Ta có: $S_{AHD}=\dfrac12AH\cdot HD=8.64$

Từ câu b $\to \dfrac{S_{AHD}}{S_{CDE}}=(\dfrac{AD}{DC})^2=\dfrac{225}{49}$

$\to S_{DEC}=\dfrac{49}{225}S_{AHD}=1.8816$