Question

In the given figure, ABD is a triangle right angled at A and $AC\perp BD$. Then,
$A{D}^2=BD\times CD$

• A. $True$
• B. $False$

Answer 1

The correct option is A $True$

In $\Delta DCAand\Delta DAB$, we have,
$\angle DCA=\angle DAB$ (Each equals to ${90}^{\circ }$)
$\angle CDA=\angle ADB$ (common angle)
$\therefore \Delta DCA\Delta DAB$ [By AA similarity criterion]
$\Rightarrow \frac{DC}{DA}=\frac{DA}{DA}$
$\Rightarrow A{D}^2=BD\times CD$