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

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

    

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In \triangle ADB\, and\, \triangle ABC,

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

\angle ABD\, =\angle CBA       (common )

\triangle ADB\, \sim \triangle ABC       (By AA)

\Rightarrow \frac{AB}{BC}=\frac{BD}{AB}

\Rightarrow AB^2=BC.BD,  hence proved

Posted by

seema garhwal

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