Question

In the figure $△ABC$ is a right angled triangle with right angle at B. BD is perpendicular to AC. Then which of the following options will hold true?

• A. $A{B}^2=AD\times DC$
• B. $A{B}^2=AD\times AC$
• C. $A{D}^2=DC\times AC$
• D. $A{B}^2=D{C}^2+A{D}^2$

Answer 1

The correct option is B

$A{B}^2=AD\times AC$

In $△ABC$ and $△ADB$
$\angle ABC=\angle ADB={90}^o$
$\angle A=\angle A$ (common angle)
Therefore, $△ABC\sim △ADB$ [by AA similarity]
$\frac{AB}{AD}=\frac{AC}{AB}$
$A{B}^2=AC\times AD$