The angle substended by a chord of a circle at the centre of the circle is equal to double the angle substended by that chord at any point on the circumference of the circle lying on a side of the chord where centre lie.

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The angle substended by a chord of a circle at the centre of the circle is equal to double the angle substended by that chord at any point on the circumference of the circle lying on a side of the chord where centre lie.

Proof :

To prove the above statement we have to consider three cases and prove it. Those are:

Case – 1:

Let the chord be line segment AB, C be the centre of the circle and P be a point on the circumference of the circle as shown in the below figure.

 

 

In the above figure, let us draw line segment CP.

Let (angle CPB) = X, then (angle CBP) = X .

Let (angle CPA) = Y, then (angle CAP) = Y.

Reason for (angle CBP) = X , (angle CAP) = Y :

If we observe the above figure,

Length of line segment CP = radius of the circle

Length of line segment CB = radius of the circle

Length of line segment CA = radius of the circle

Therefore,

As the length of line segment CP = Length of line segment CB, (angle CPB) = (angle CBP).

Similarly ,

As the length of line segment CP = Length of line segment CA, (angle CPA) = (angle CAP).

Therefore,

(angle CBP) = X , (angle CAP) = Y.

Then the diagram will be as shown in the below figure.

 

Therefore, the angle substended by the chord AB at the point P is equal to (X + Y).

Then in triangle CPB,

X + X + (angle PCB) = 180

(Angle PCB) = 180 – 2X

Similarly,

From triangle PCA,

Angle PCA = 180 – 2Y

At point C,

(Angle ACB) + (Angle PCB) + (angle PCA) = 360

From above data,

(Angle ACB) + (180 – 2X) + (180 – 2Y) = 360

Then , (angle ACB) = 2X + 2Y

i.e (angle ACB) = 2(X + Y)

Here (angle ACB) is the angle substended by the chord AB at the centre of the circle.

From above data,

Therefore,

The angle substended by a chord of a circle at the centre of the circle is equal to double the angle substended by that chord at any point on the circumference of the circle lying on a side of the chord where centre lie.


Case – 2:

Let the chord be line segment AB, C be the centre of the circle and P be a point on the circumference of the circle as shown in the below figure.


In the above figure,

Let (angle CPB) = X, then (angle PBC) = X

Reason for this is same as reason in case – 1.

Therefore, the angle substended by the chord AB at the point P is equal to X.

Then in triangle PCB,

(angle PCB) + (angle CPB) + (angle PBC) = 180

From above data,

(angle PCB) + X + X = 180

Therefore, (angle PCB) = 180 – 2X

As a line segment is joining the points A, C and P,

(angle ACB) + (angle PCB) = 180

From above information,

Then , (angle ACB) + (180 – 2X) = 180

Therefore, (angle ACB) = 2X

Here (angle ACB) is the angle substended by the chord AB at the centre of the circle.

From above data,

Therefore,

The angle substended by a chord of a circle at the centre of the circle is equal to double the angle substended by that chord at any point on the circumference of the circle lying on a side of the chord where centre lie.

Case – 3:

Le the chord be line segment AB, C be the centre of the circle and P be a point on the circumference of the circle as shown in the below figure.

 

 

In the above figure ,let us join C and P.

Let (angle APB) = X and (angle CPA) = Y

Then, (angle CAP) = Y

Reason for this is same as reason in case – 1.

Therefore, the angle substended by the chord AB at the point P is equal to X.

Then the diagram will be as shown in the below figure.

From above figure, (angle CPB) = X + Y , then

(angle PBC) = X + Y

Reason for this is same as the reason in case - 1.

From triangle BCP,

(angle BCP) + (angle CPB) + (angle PBC) = 180

From above information,

(angle BCP) + (X + Y) + (X + Y) = 180

Then, (angle BCP) = 180 – 2X – 2Y

Similarly,

From triangle ACP ,

(Angle ACP) = 180 – 2Y

Here ,

(angle ACB) = (angle ACP) – (angle BCP)

From above data,

(angle ACB) = (180 – 2Y) – (180 – 2X – 2Y)

Therefore, (angle ACB) = 2X

Here (angle ACB) is the angle substended by the chord AB at the centre of the circle.

From above data,

Therefore,

The angle substended by a chord of a circle at the centre of the circle is equal to double the angle substended by that chord at any point on the circumference of the circle lying on a side of the chord where centre lie.

If we consider any other point on the circumference of the circle which is on a side of the chord where centre lie and try to prove, it comes under any of the above three situations and then it will also get proved.

Therefore,

The angle substended by a chord of a circle at the centre of the circle is equal to double the angle substended by that chord at any point on the circumference of the circle lying on a side of the chord where centre lie.


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