a,
$\Delta$ ABC và $\Delta$ HBA có:
$\widehat{BAC}=\widehat{BHA}= 90^o$
$\widehat{ABC}$ chung
=> $\Delta$ ABC $\backsim$ $\Delta$ HBA (g.g) (1)
b,
CMTT câu a, ta có $\Delta$ ABC $\backsim$ $\Delta$ HCA (g.g) (2)
(1)(2) => $\Delta$ HBA $\backsim$ $\Delta$ HAC
=> $\frac{AH}{HB}=\frac{HC}{AH}$
$\Leftrightarrow AH^2=HB.HC$
c,
$\Delta$ ABC vuông tại A có $BC=\sqrt{AB^2+AC^2}= 10cm$
(1) => $\frac{AH}{AB}=\frac{AC}{BC}$
$\Leftrightarrow AH=\frac{AB.AC}{BC}=\frac{6.8}{10}=4,8cm$
d,
(2) => $\frac{AC}{CH}=\frac{BC}{AC}$
$\Leftrightarrow HC=\frac{AC^2}{BC}= 6,4cm$
$\Delta$ ACD và $\Delta$ HCE có:
$\widehat{DAC}=\widehat{EHC}=90^o$
$\widehat{ACD}=\widehat{HCE}$
=> $\Delta$ ACD $\backsim$ $\Delta$ HCE
=> $k=\frac{AC}{HC}=\frac{8}{6,4}= \frac{5}{4}$
=> $\frac{S_{ACD}}{S_{HBE}}= k^2= \frac{25}{16}$
