Question

In the given figure, a square is inscribed in a circle with center O. Find:

1. ∠BOC
2. ∠OCB
3. ∠COD
4. ∠BOD

Is BD a diameter of the circle?

Answer 1

In the given figure we can extend the straight line OB to BD and CO to CA

Then we get the diagonals of the square which intersect each other at 90 by the property of Square.

From the above statement, we can see that
∠COD = 90°

The sum of the angle ∠BOC and ∠COD is 180° as BD is a straight line.
Hence ∠BOC + ∠OCD = ∠BOD = 180°
∠BOC + 90° = 180°
∠BOC + 180° - 90°
∠BOC = 90°

We can see that the OCB is an isosceles triangle with sides OB and OC of EQual length as they are the radii of the same are.

In ΔOCB,
∠OBC = ∠OCB as they are opposite angles to the two equal sides of an isosceles triangle.

Sum of all the angles of a triangle is 180°
so, ∠OBC + ∠OCB + ∠BOC =180°
∠OBC + ∠OBC + 90° = 180° as, ∠OBC = ∠OCB
2∠OBC = 180° - 90°
2∠OBC = 90°
2∠OBC = 45°
as ∠OBC = ∠OCB So,
∠OBC = OCB = 45°

Yes BD is the diameter of the order.

Answer 2

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° by the property of square.

From the above statement, we can see that
∠COD = 90°.

The sum of the angle ∠BOC and ∠OCD is 180° as BD is a straight line.

Hence ∠BOC + ∠OCD = ∠BOD = 180°

∠BOC + 90° = 180°

∠BOC + 180° - 90°

∠BOC = 90°

We can see that the OCB is an isosceles triangle with sides OB and OC of EQual length as they are the radii of the same circle.

In ΔOCB,

∠OBC = ∠OCB    ...[Opposite angles to the two equal sides of an isosceles triangle]

Sum of all the angles of a triangle is 180°.

so, ∠OBC + ∠OCB + ∠BOC =180°

∠OBC + ∠OBC + 90° = 180°    ...[∠OBC = ∠OCB]

2∠OBC = 180° - 90°

2∠OBC = 90°

2∠OBC = 45°

as ∠OBC = ∠OCB

So, ∠OBC = ∠OCB = 45°

Yes, BD is the diameter of the order.