Đáp án : $b.BC=20, AH=\dfrac{48}{5}, HB=\dfrac{36}5$
$c.DB=\dfrac{60}7, DC=\dfrac{80}7$
Giải thích các bước giải:
a.Ta có : $AH\perp BC\to\widehat{AHB}=\widehat{BAC}\to\Delta HBA\sim\Delta ABC(g.g)$
b.Vì $\hat A=90^o\to BC^2=AB^2+AC^2=400\to BC=20$
Ta có : $AH\perp BC\to AH.BC=AB.AC=2S_{ABC}$
$\to AH=\dfrac{AB.AC}{BC}=\dfrac{48}{5}$
$\to HB=\sqrt{AB^2-AH^2}=\dfrac{36}{5}$
c.Vì AD là phân giác góc A
$\to\dfrac{DB}{DC}=\dfrac{AB}{AC}=\dfrac{3}{4}$
$\to\dfrac{DB}{DB+DC}=\dfrac{3}{3+4}\to\dfrac{BD}{BC}=\dfrac{3}{7}$
$\to BD=\dfrac37BC=\dfrac{60}{7}$
$\to CD=BC-BD=\dfrac{80}{7}$