Question

Question 3 (iii)
In the given figure, ABD is a triangle right angled at A and $AC\perp BD$. Show that
$\left(iii\right)A{D}^2=BD\times CD$

Answer 1

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