ABC is a triangle right angled at B and $\text{BD}\perp\text{AC}.$ If AD = 4cm, and CD = 5cm, find BD and AB.
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In $\triangle\text{ABC},$$\angle\text{ABC}=90^\circ$ [Given]
$\text{BD}\perp\text{AC}$ [Hypotenuse]
$\therefore\text{BD}^2=\text{DA}\times\text{DC}$
$\Rightarrow\text{BD}^2=4\times5$
$\Rightarrow\text{BD}=2\sqrt{5}\text{cm}$
In right angled $\triangle\text{BDA},$
$\text{BD}\perp\text{AC}$ [Given]
$\therefore\angle\text{BDA}=90^\circ$
$\Rightarrow\text{AB}^2=\text{AD}^2+\text{BD}^2$ [by Pythagoras theorem]
$\Rightarrow\text{AB}^2=4^2+(2\sqrt{5})^2$
$\Rightarrow\text{AB}^2=16+20$
$\Rightarrow\text{AB}^2=36$
$\Rightarrow\text{AB}=6\text{cm}$
Hence proved.
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