Giải thích các bước giải:
a.Ta có: $HA\perp BC\to \widehat{AHB}=\widehat{BAC}=90^o$
$\to \Delta HBA\sim\Delta ABC(g.g)$
b.Ta có : $BC^2=AB^2+AC^2=400\to BC=20$
Mà $AH\perp BC, AB\perp AC\to AH.BC=AB.AC=2S_{ABC}$
$\to AH=\dfrac{AB.AC}{BC}=\dfrac{48}5$
$\to BH=\sqrt{AB^2-AH^2}=\dfrac{36}5$
c.Vì AD là phân giác góc A
$\to \dfrac{DB}{DC}=\dfrac{BA}{AC}=\dfrac34$
$\to \dfrac{DB}{BD+DC}=\dfrac{3}{3+4}$
$\to \dfrac{DB}{BC}=\dfrac37$
$\to BD=\dfrac37BC=\dfrac{60}7\to DC=BC-BD=\dfrac{80}7$
d.Ta có : $MN//BC\to \Delta AMN\sim\Delta ABC(g.g)$
$\to \dfrac{S_{AMN}}{S_{ABC}}=(\dfrac{AM}{AB})^2=(\dfrac{AK}{AH})^2=(\dfrac38)^2=\dfrac{9}{64}$
$\to \dfrac{S_{ABC}-S_{AMN}}{S_{ABC}}=\dfrac{64-9}{64}$
$\to \dfrac{S_{BCMN}}{S_{ABC}}=\dfrac{55}{64}$
$\to S_{BMNC}=\dfrac{55}{64}S_{ABC}=\dfrac{165}2$
