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  • In Fig. 6.53, ABD is a triangle right angled at A and AC perpendicular BD. Show that AD ^ 2 = DC . CD

Q3 In Fig. 6.53, ABD is a triangle right angled at A and AC \perp  BD. Show that  AD^2 = BD . CD .

  

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In \triangle ACD\, and\, \triangle ABD,

\angle DCA\, =\angle DAB\, \, \, \, \, \, \, \, (Each 90 \degree)

\angle CDA\, =\angle ADB       (common )

\triangle ACD\, \sim \triangle ABD       (By AA)

\Rightarrow \frac{CD}{AD}=\frac{AD}{BD}

\Rightarrow AD^2=BD\times CD

Hence proved

Posted by

seema garhwal

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