Question

In the given figure , if $$ \angle BAC = 90^{\circ} $$ and $$ AD \perp BC $$ . Then ,

Answer 1

Correct option is C. $$ BD . CD = AD^{2}$$
In $$ \Delta ADB $$ and $$ \Delta ADC $$ ,
$$ \angle \, BDA = \angle \, ADC = 90^{\circ} $$ [Given]
$$ \angle B = \angle DAC = (90^{\circ} - C) $$
$$ \therefore \, \Delta ADB \sim \Delta CDA $$ [ By AA similarity criterion]
$$ \Rightarrow \, \dfrac{AD}{CD} = \dfrac{AB}{CA} = \dfrac{DB}{DA} $$
$$ \therefore \, AD^2 = BD \cdot CD $$