Q

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

Q3 (1)   In Fig. 6.53, ABD is a triangle right angled at A and AC $\perp$  BD. Show that
$AB^2 = BC . BD .$

Views

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 prooved

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