Question
In the given figure, $\angle\text{CAB}=90^\circ$ and $\text{AD}\perp\text{BC}.$ Show that $\triangle\text{BDA}\sim\triangle\text{BAC}.$ If $AC = 75\ cm, AB = 1m,$ and $BC = 1,25m$ find $AD.$
✓
Answer
Given: $AB = 100\ cm, BC = 125\ cm, AC = 75\ cm$
Proof:
In $\triangle\text{BAC}$ and $\triangle\text{BDA}$
$\angle\text{BAC}=\angle\text{BDA}=90^\circ$
$\angle\text{B}=\angle\text{B}($ common$)$
$\triangle\text{BAC}\sim\triangle\text{BDA}($ by $AA$ similarities$)$
$\Rightarrow\frac{\text{BA}}{\text{BC}}=\frac{\text{AD}}{\text{AC}}$
$\Rightarrow\frac{\text{100}}{\text{125}}=\frac{\text{AD}}{\text{75}}$
$\Rightarrow\text{AD}=\frac{\text{100}\times75}{\text{125}}=60\text{ cm}$
Therefore, $AD = 60\ cm$
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