Giải thích các bước giải:

a.Ta có: $AH\perp BC\to\widehat{AHB}=\widehat{BAC}=90^o$

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

b.Từ câu a$\to \widehat{BAH}=\widehat{ACB}=\widehat{ACH}$

Mà $\widehat{AHB}=\widehat{AHC}$

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

$\to \dfrac{AH}{CH}=\dfrac{BH}{AH}\to AH^2=HB.HC$

c.Ta có : $AB\perp AC\to BC^2=AB^2+AC^2=100\to BC=10$

Mà $AB.AC=AH.BC(=2S_{ABC})$

$\to AH=\dfrac{AB.AC}{BC}=\dfrac{24}5$

d.Ta có : $CH^2=AC^2-AH^2=8^2-(\dfrac{24}5)^2=\dfrac{1024}{25}$

$\to CH=\dfrac{32}{5}$

Vì CD là phân giác góc C

$\to\widehat{ACD}=\widehat{ECH}$

Mà $\widehat{DAC}=\widehat{EHC}=90^o$

$\to \Delta CAD\sim\Delta CHE(g.g)$

$\to \dfrac{S_{ACD}}{S_{HCE}}=(\dfrac{CA}{CH})^2=(\dfrac{8}{\dfrac{32}{5}})^2=\dfrac{25}{16}$