Question
In the given figure, a square is inscribed in a circle with center $O$. Find:
- $∠BOC$
- $∠OCB$
- $∠COD$
- $∠BOD$
Is $BD$ a diameter of the circle?
✓
Answer
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
$\angle COD = 90^\circ $
The sum of the angle$ \angle BOC$ and $\angle COD$ is $180^\circ $ as $BD$ is a straight line.
Hence $\angle BOC + \angle OCD = \angle BOD = 180^\circ $
$\angle BOC + 90^\circ = 180^\circ $
$\angle BOC + 180^\circ - 90^\circ $
$\angle BOC = 90^\circ $
We can see that the $\text{OCB}$ is an isosceles triangle with sides $OB$ and $OC$ of EQual length as they are the radii of the same are.
In $\triangle OCB,$
$\angle OBC = \angle OCB$ as they are opposite angles to the two equal sides of an isosceles triangle.Sum of all the angles of a triangle is $180^\circ $
so, $\angle OBC + \angle OCB + \angle BOC =180^\circ $
$\angle OBC + \angle OBC + 90^\circ = 180^\circ $ as, $\angle OBC = \angle OCB$
$2\angle OBC = 180^\circ - 90^\circ $
$2\angle OBC = 90^\circ $
$2\angle OBC = 45^\circ $
as $\angle OBC = \angle OCB$ So,
$\angle OBC = \text{OCB} = 45^\circ $
Yes $BD$ is the diameter of the order.
From the above statement, we can see that
$\angle COD = 90^\circ $
The sum of the angle$ \angle BOC$ and $\angle COD$ is $180^\circ $ as $BD$ is a straight line.
Hence $\angle BOC + \angle OCD = \angle BOD = 180^\circ $
$\angle BOC + 90^\circ = 180^\circ $
$\angle BOC + 180^\circ - 90^\circ $
$\angle BOC = 90^\circ $
We can see that the $\text{OCB}$ is an isosceles triangle with sides $OB$ and $OC$ of EQual length as they are the radii of the same are.
In $\triangle OCB,$
$\angle OBC = \angle OCB$ as they are opposite angles to the two equal sides of an isosceles triangle.Sum of all the angles of a triangle is $180^\circ $
so, $\angle OBC + \angle OCB + \angle BOC =180^\circ $
$\angle OBC + \angle OBC + 90^\circ = 180^\circ $ as, $\angle OBC = \angle OCB$
$2\angle OBC = 180^\circ - 90^\circ $
$2\angle OBC = 90^\circ $
$2\angle OBC = 45^\circ $
as $\angle OBC = \angle OCB$ So,
$\angle OBC = \text{OCB} = 45^\circ $
Yes $BD$ is the diameter of the order.
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