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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)$$\triangle AHD \sim \triangle BHA(g.g)$Vì:$\widehat{DAH}=\widehat{ABD}$(cùng cộng với$\widehat{ADB}=90^o$)$\widehat{AHD}=\widehat{AHB}=90^o$$\Rightarrow \frac{AB}{BD}=\frac{AH}{AD}\Rightarrow AH=\frac{AB.AD}{BD}=\frac{12.9}{15}=7,2$ $(cm)$c) $S_{ABC}=\frac{1}{2}.AB.BC=54$ $(cm^2)$

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) $S_{ABC}=\frac{1}{2}.AB.BC=54$ $(cm^2)$

a) $\triangle AHB\sim \triangle BCD$ $(g.g)$.Tự chứng minh.b) $\triangle AHD \sim \triangle BHA$ $(g.g)$$BD=\sqrt{BC^2+AB^2}=15$. $(cm)$$\Rightarrow \frac{AB}{BD}=\frac{AH}{AD}\Rightarrow AH=\frac{AB.AD}{BD}=\frac{12.9}{15}=7,2$ $(cm)$c) $S_{ABC}=\frac{1}{2}.AB.BC=54$ $(cm^2)$

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)$$\triangle AHD \sim \triangle BHA(g.g)$Vì: $\widehat{DAH}=\widehat{ABD}$ (cùng cộng với $\widehat{ADB}=90^o$)$\widehat{AHD}=\widehat{AHB}=90^o$$\Rightarrow \frac{AB}{BD}=\frac{AH}{AD}\Rightarrow AH=\frac{AB.AD}{BD}=\frac{12.9}{15}=7,2$ $(cm)$c) $S_{ABC}=\frac{1}{2}.AB.BC=54$ $(cm^2)$