1/
a,
$\Delta$ AHB và $\Delta$ BCD có:
$\widehat{AHB}= \widehat{BCD}= 90^o$
$\widehat{ABH}= \widehat{BDC}$ (SLT)
=> $\Delta$ AHB $\backsim$ $\Delta$ BCD (g.g) (*)
b,
$\Delta$ BCD vuông tại C có BD= $\sqrt{9^2+12^2}= 15cm$
(*) => $\frac{AH}{BC}= \frac{AB}{BD}= \frac{BH}{CD}$
=> $\frac{AH}{9}= \frac{12}{15}= \frac{BH}{12}$
=> $AH= 7,2; BH= 9,6$
=> $S_{AHB}= 0,5AH.BH= 34,56 (cm^2)$
2/
$\widehat{HAB} = \widehat{BAC}-\widehat{HAC}= 90^o - \widehat{HAC}$(*)
$\widehat{HCA}= 90^o - \widehat{HAC}$(**)
(*)(**) => $\widehat{HAB}= \widehat{HCA}$ (1)
Mà $\widehat{BHA}= \widehat{AHC}= 90^o$ (2)
(1)(2)=> $\Delta$ BHA $\backsim$ $\Delta$ AHC (g.g)
=> $\frac{BH}{AH}= \frac{AH}{HC}$
=> $AH^2= BH.HC= 9.16= 144$
=> $AH= 12$
$BC= 9+16= 25$
Theo Pytago:
$AB= \sqrt{AH^2+ BH^2}= 15$
$AC= \sqrt{AH^2+ HC^2}= 20$
