Giải thích các bước giải:

a.Ta có: $AD=6, AB=8$

$\to BD=\sqrt{AD^2+AB^2}=10$

Vì $AI$ là phân giác $\widehat{DAB}$

$\to \dfrac{ID}{IB}=\dfrac{AD}{AB}=\dfrac34$

$\to \dfrac{ID}3=\dfrac{IB}4=\dfrac{ID+IB}{3+4}=\dfrac{BD}7=\dfrac{10}7$

$\to ID=\dfrac{30}7, IB=\dfrac{40}7$

b.Xét $\Delta HBA, \Delta ABD$ có:

Chung $\hat B$

$\hat H=\hat A(=90^o)$

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

c.Xét $\Delta AHD, \Delta AHB$ có:

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

$\widehat{HAD}=90^o-\widehat{HAB}=\widehat{HBA}$

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

$\to \dfrac{HA}{HB}=\dfrac{HD}{HA}$

$\to AH^2=HD.HB$

d.Ta có: $\Delta ABD$ vuông tại $A, AH\perp BD$

$\to S_{ABD}=\dfrac12AH\cdot BD=\dfrac12AB\cdot AD$

$\to AH=\dfrac{AB\cdot AD}{DB}=4.8$