Giải thích các bước giải:
a.Ta có : $AH\perp BC\to \widehat{AHB}=\widehat{BAC}=90^o$
$\to \Delta ABH\sim\Delta CBA(g.g)$
b.Ta có $\hat A=90^o$
$\to BC^2=AB^2+AC^2=100\to BC=10$
Mà $AH\perp BC\to S_{ABC}=\dfrac12AH.BC=\dfrac12AB.AC$
$\to AH=\dfrac{AB.AC}{BC}=\dfrac{24}5$
$\to BH^2=AB^2-AH^2= \dfrac{324}{25}$
$\to BH=\dfrac{18}{5}$
c.Vì $AD$ là phân giác góc $A$
$\to \dfrac{DB}{DC}=\dfrac{AB}{AC}=\dfrac34$
$\to \dfrac{DB}{DB+DC}=\dfrac3{3+4}$
$\to \dfrac{DB}{BC}=\dfrac37$
$\to DB=\dfrac37BC=\dfrac{30}{7}$
$\to DC=BC-BD=\dfrac{40}7$
d.Ta có : $MN//BC$
$\to \widehat{AMN}=\widehat{ABC},\widehat{ANM}=\widehat{ACB}$
$\to \Delta AMN\sim\Delta ABC(g.g)$
$\to \dfrac{S_{AMN}}{S_{ABC}}=(\dfrac{AM}{AB})^2$
Mà $MN//BC\to \dfrac{AM}{AB}=\dfrac{AK}{AH}=\dfrac{1.8}{\dfrac{24}{5}}=\dfrac38$
$\to \dfrac{S_{AMN}}{S_{ABC}}=\dfrac{9}{64}$
$\to \dfrac{S_{BCMN}}{S_{ABC}}=\dfrac{55}{64}$
$\to S_{BCMN}=\dfrac{55}{64}S_{ABC}$
$\to S_{BCMN}=\dfrac{55}{64}\cdot\dfrac12\cdot AB\cdot AC$
$\to S_{BCMN}=\dfrac{55}{64}\cdot\dfrac12\cdot 6\cdot 8$
$\to S_{BCMN}=\dfrac{165}{8}$
