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 = 75cm, AB = 1m, and BC = 1,25m find AD.
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Given: AB = 100cm, BC = 125cm, AC = 75cm
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 = 60cm
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