Question

Question 1
In figure, if $\angle BAC={90}^{\circ }andAD\perp BC$. Then,


(A) $BD.CD=B{C}^2$
(B) $AB.AC=B{C}^2$
(C) $BD.CD=A{D}^2$
(D) $AB.AC=A{D}^2$

Answer 1

(C) $BD.CD=A{D}^2$

$In\Delta ADBand\Delta ADC,$

$\angle ADB=\angle ADC={90}^{\circ }$
$\angle DBA=\angle DAC$ [ each equal to ${90}^{\circ }$ - < C]
$\therefore$ $\Delta ADB\sim \Delta ADC$ [ by AAA similarity criterion ]
$\therefore \frac{BD}{AD}=\frac{AD}{CD}$
$\Rightarrow BD.CD=A{D}^2$