Question

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

Answer 1

(i) In $\Delta ADB$ and $\Delta CAB$, we have
$\angle DAB=\angle ACB$ (Each angle is equal to ${90}^{\circ }$)
$\angle ABD=\angle CBA$ (Common angle)
$\therefore \Delta ADB\sim \Delta CAB$ [AAA similarity criterion]
$\Rightarrow \frac{AB}{CB}=\frac{BD}{AB}$
$\Rightarrow A{B}^2=CB\times BD$
Therefore,$A{B}^2=BC\times BD$