Question

In the given figure, a square is inscribed in a circle with centre O. Find:
(i) ∠ BOC
(ii) ∠ OCB
(iii) ∠ COD
(iv) ∠ BOD
Is BD a diameter of the circle?

Answer 1

From the figure, extend a straight-line OB to BD and CO to CA.
We get the diagonals of the square which intersect each other at ${90}^o$ by the property of square.



From the above mentioned statement, we know that
$\angle COD={90}^o$
Here, the sum of the angle $\angle BOCand\angle OCD$ is ${180}^o$ as BD is a straight line.

$\angle BOC+\angle OCD=\angle BOD={180}^o$

It can be written as

$\angle BOC+{90}^o={180}^o$

$\angle BOC={180}^o–{90}^o$

$\angle BOC={90}^o$

Therefore, triangle OCB is an isosceles triangle with sides OB and OC of equal length as they are the radii of the same circle.

In $\Delta$ OCB,
$\angle OBC=\angle OCB$ [Opposite angles to the two equal sides of an isosceles triangle]

Here sum of all the angles of a triangle is ${180}^o$

$\angle OBC+\angle OCB+\angle BOC={180}^o$

It can be written as

$\angle OBC+\angle OBC+{90}^o={180}^o[\angle OBC=\angle OCB]$

So we get

$2\angle OBC={180}^o–{90}^o$

$2\angle OBC={90}^o$

$\angle OBC={45}^o$

Here, $\angle OBC=\angle OCB={45}^o$

Yes, BD is the diameter of the circle.