Giải thích các bước giải:

a.Ta có $\Delta ABC$ vuông tại $A$

$\to BC^2=AB^2+AC^2=100$

$\to BC=10$

b.Xét $\Delta ABC,\Delta HBA$ có:

Chung $\hat B$

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

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

$\to \widehat{HAB}=\widehat{ACB}=\widehat{ACH}$

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

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

c.Ta có $M, N$ là trung điểm $HB, HA$

$\to MN$ là đường trung bình $\Delta HAB$

$\to MN//AB$

Mà $AB\perp AC\to MN\perp AC$

Do $AH\perp BC\to AN\perp CM$

$\to N$ là trực tâm $\Delta ACM$

$\to CN\perp AM$

d.Xét $\Delta BAK,\Delta BAD$ có:
Chung $\hat B$

$\widehat{BKA}=\widehat{BAD}(=90^o)$

$\to\Delta BAK\sim\Delta BDA(g.g)$

$\to\dfrac{BA}{BD}=\dfrac{BK}{BA}$

$\to BA^2=BK\cdot BD$

Từ câu a $\to \dfrac{BA}{BC}=\dfrac{BH}{BA}$

$\to BA^2=BH\cdot BC$

$\to BK\cdot BD=BH\cdot BC$

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

Do $\widehat{HBK}=\widehat{CBD}$

$\to\Delta KBH\sim\Delta CBD(c.g.c)$

e.Ta có: $CD=AD+AC=9$

$\to S_{BCD}=\dfrac12AB\cdot CD=36$

Vì $\Delta BKH\sim\Delta BCD$(câu d)

$\to \dfrac{S_{BKH}}{S_{BCD}}=(\dfrac{BH}{BD})^2$

Ta có: $BD^2=AD^2+AB^2=73$

            $AH\perp BC\to S_{ABC}=\dfrac12AB\cdot AC=\dfrac12AH\cdot BC\to AH=\dfrac{AB\cdot AC}{BC}=4.8$

$\to BH^2=AB^2-AH^2=40.96$

$\to \dfrac{S_{BKH}}{S_{BCD}}=\dfrac{1024}{1825}$

$\to S_{BKH}\approx 20.20$