a,
$\Delta$ ABC và $\Delta$ HBA có:
$\widehat{ABC}$ chung
$\widehat{BAC}=\widehat{BHA}=90^o$
$\Rightarrow \Delta$ ABC $\backsim$ $\Delta$ HBA (g.g) (*)
b,
$\Delta$ AHB và $\Delta$ CHA có:
$\widehat{AHB}=\widehat{CHA}=90^o$
$\widehat{BAH}=\widehat{ACH}=90^o-\widehat{HAC}$
$\Rightarrow \Delta$ AHB $\backsim$ $\Delta$ CHA (g.g)
$\Rightarrow \frac{AH}{BH}=\frac{HC}{AH}$
$\Leftrightarrow AH^2=BH.HC$
c,
$BC=\sqrt{AB^2+AC^2}=10cm$
(*) $\Rightarrow \frac{AH}{AC}=\frac{AB}{BC}$
$\Rightarrow AH=4,8cm$
d,
$HC=\sqrt{AC^2-AH^2}=6,4cm$
$\Delta$ ADC và $\Delta$ HEC có:
$\widehat{DAC}=\widehat{EHC}=90^o$
$\widehat{ACD}=\widehat{ECH}$
$\Rightarrow \Delta$ ADC $\backsim$ $\Delta$ HEC (g.g)
$\Rightarrow \frac{S_{ADC}}{S_{HEC}}=k^2= (\frac{AC}{HC})^2=\frac{25}{16}$