Giải thích các bước giải:

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

Vì BD là phân giác góc B
$\to \dfrac{DA}{DC}=\dfrac{BA}{BC}=\dfrac35$

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

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

$\to AD=\dfrac38AC=3$

$\to CD=AC-AD=5$

b.Ta có : $AH\perp BC\to AH.BC=AB.AC(=2S_{ABC})$

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

$\to BH=\sqrt{AB^2-AH^2}=\dfrac{18}5$

$\to CH=BC-BH=\dfrac{32}5$

c.Ta có : $\widehat{AHB}=\widehat{BAC}=90^o$

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

$\to \dfrac{AB}{CB}=\dfrac{BH}{AB}\to AB^2=BH.BC$

d.Ta có : $\widehat{BMA}=\widehat{CEA}=90^o(BM,CE\perp xy)$

Mà $\widehat{MAB}+\widehat{EAC}=180^o-\widehat{BAC}=90^o=\widehat{EAC}+\widehat{ECA}$

$\to \widehat{MAB}=\widehat{ECA}$

$\to \Delta ABM\sim\Delta CAE(g.g)$

$\to \dfrac{S_{AMB}}{S_{CNA}}=(\dfrac{AB}{AC})^2=\dfrac9{16}$