Giải thích các bước giải:

 a.Ta có $\widehat{ADI}=\widehat{IHC}=90^o,\widehat{AID}=\widehat{HIC}$

$\to\Delta AID\sim\Delta CIH(g.g)$

b.Ta có: $\widehat{BHA}=\widehat{BDC}=90^o,\widehat{DBC}=\widehat{ABH}$

$\to\Delta BAH\sim\Delta BCD(g.g)$

$\to\dfrac{BA}{BC}=\dfrac{BH}{BD}$

$\to BA.BD=BH.BC$

c.Ta có :

$AH\perp BC\to AH=\sqrt{AB^2-BH^2}=4$

$\to CH=\sqrt{AC^2-AH^2}=\sqrt{9^2-4^2}=\sqrt{65}$

$\to BC=BH+CH=3+\sqrt{65}$

$\to S_{ABC}=\dfrac12\cdot AH\cdot BC=\dfrac12\cdot 4\cdot (3+\sqrt{65})=2(3+\sqrt{65})$

d.Từ câu b

$\to\dfrac{BD}{BC}=\dfrac{BH}{BA}$

Mà $\widehat{DBH}=\widehat{ABC}$

$\to \Delta BHD\sim\Delta BAC(c.g.c)$

e.Ta có $BD\cdot BA=BH\cdot BC$

$\to BD=\dfrac{BH\cdot BC}{BA}=\dfrac{3\cdot (3+\sqrt{65})}{5}$

Từ câu d

$\to\dfrac{DH}{AC}=\dfrac{BH}{BA}$

$\to DH=\dfrac{BH\cdot AC}{BA}=\dfrac{3\cdot 9}{5}=\dfrac{27}5$