Question

In a right angled triangle $\angle A$ is a right-angle and AO is perpendicular to BC at point O, then ${AO}^2=BO\times CO$ .

• A. True
• B. False

Answer 1

The correct option is A

True

Let in $\Delta ABC$, $\angle A$ is a right angle.

AO is perpendicular to BC at O $\therefore$ $\angle AOC$ is a right angle.

Now in $\Delta AOB$ and $\Delta AOC$

$\angle AOB=\angle AOC={90}^{\circ }$

In $\Delta ABC$,

$\angle ABC+\angle ACB={90}^{\circ }$

$\Rightarrow \angle ABC={90}^{\circ }-\angle ACB={90}^{\circ }-\angle ACO\left(1\right)$

In $\Delta ACO$,

$\angle OAC+\angle ACO={90}^{\circ }$

$\Rightarrow \angle OAC={90}^{\circ }-\angle ACO\left(2\right)$

From (1) and (2) we get:

$\angle OAC=\angle ABC=\angle ABO$

So, $\Delta AOB\sim \Delta COA$

$\therefore \frac{AO}{CO}=\frac{BO}{AO}$

$\Rightarrow {AO}^2=BO\times CO$