a,
$\Delta$ ABC vuông tại A.
$\Rightarrow BC=\sqrt{AB^2+AC^2}=10cm$
AD phân giác $\Rightarrow \frac{BD}{AB}=\frac{DC}{AC}=\frac{BD+DC}{AB+AC}=\frac{10}{6+8}=\frac{5}{7}$
$\Rightarrow BD=\frac{5}{7}.AB=\frac{30}{7}cm$
b,
$\Delta$ AHB và $\Delta$ CHA có:
$\widehat{BAH}=\widehat{ACH}=90^o-\widehat{HAC}$
$\widehat{BHA}=\widehat{AHC}=90^o$
$\Rightarrow \Delta$ AHB $\backsim$ $\Delta$ CHA (g.g)
$\Rightarrow \frac{AH}{HB}=\frac{HC}{AH}$
$\Leftrightarrow AH^2=HB.HC$
c,
ED // AC ($\bot$ AB)
$\Rightarrow \frac{EB}{AB}=\frac{BD}{BC}$
$\Rightarrow AB.BD=EB.BC$
d,
$\Delta$ AHB $\backsim$ $\Delta$ CAB (g.g)
$\Rightarrow \frac{AB}{BH}=\frac{BC}{AB}$
$\Leftrightarrow BH=\frac{AB^2}{BC}=3,6cm$
$\Rightarrow HC=BC-HB=6,4cm$
$AH=\sqrt{BH.HC}=4,8cm$
$HD= BD-BH=\frac{24}{35}cm$
$\Rightarrow S_{AHD}=\frac{1}{2}AH.HD\approx 1,6(cm^2)$
