Question

In the following figure , OA = OC and AB = BC ,
prove that :
$$ \angle AOB = 90^{o}$$

Answer 1

In the figure OA = OC , AB = BC

We need to proved that $$ \angle AOB = 90^{o}$$

In $$ \Delta ABO$$ and $$\Delta CBO $$

$$AB = BC $$ [given ]

$$OB = OB$$ [common]

$$OA=OC$$ [given]

By Side - Side - Side criterion of congruence , we have
$$ \Delta ABO \cong \Delta CBO $$
The corresponding parts of the congruent triangles are congruent

$$ \Rightarrow \angle AOB = \angle COB$$ [C.P.C,T] ...(1)

Also,

$$\angle AOB + \angle COB=180^0$$ [Supplementary angles]

Using (1), we can write:

$$\angle AOB + \angle AOB=180^0$$

$$\Rightarrow 2\angle AOB =180^0$$

$$\Rightarrow \angle AOB =90^0$$