a) $\triangle AHB\sim \triangle BCD$ $(g.g)$Vì: $\widehat{ABH}=\widehat{BDC}$ (sole trong)
$\widehat{AHB}=\widehat{BCD}=90^o$.
b) $BD=\sqrt{BC^2+AB^2}=15$. $(cm)$
$\triangle AHD \sim \triangle BAD$ $(g.g)$
Vì: $\widehat{DAH}=\widehat{ABD}$ (cùng cộng với $\widehat{ADB}=90^o$)
$\widehat{AHD}=\widehat{DAB}=90^o$
$\Rightarrow \frac{AB}{BD}=\frac{AH}{AD}\Rightarrow AH=\frac{AB.AD}{BD}=\frac{12.9}{15}=7,2$ $(cm)$
c) $S_{ABC}=\frac{1}{2}.AB.BC=54$ $(cm^2)$