Question

(Prove) The angle subtended by an arc at the center is double the angle subtended by it at any

Answer 1

Answer:

The angle subtended by an arc at the center is double the angle subtended by it at any  point on the remaining part of the circle.

Step-by-step explanation:

Let us consider the image attached.

Center of circle be O.

Arc AB subtends the angle $\angle APB$ on the circle and $\angle AOB$ on the center of the circle.

To prove:

$\angle AOB = 2 \times \angle APB$

Proof:

In $\triangle PAO$: AO and PO are radius of the circles so AO = PO

And angles opposite to equal sides of a triangle are also equal in a triangle.

So, $\angle PAO = \angle OPA$

Using external angle property, that external angle is equal to sum of two opposite internal angles of a triangle.

$\angle AOQ = \angle PAO + \angle OPA=2 \times \angle APO .... (1)$

Similarly,

In $\triangle PBO$: BO and PO are radius of the circles so BO = PO

And angles opposite to equal sides of a triangle are also equal in a triangle.

So, $\angle PBO = \angle OPB$

Using external angle property, that external angle is equal to sum of two opposite internal angles of a triangle.

$\angle BOQ = \angle PBO + \angle OPB=2 \times \angle BPO .... (2)$

Now, we can see that:

$\angle AOB = \angle AOQ+\angle BOQ$

Using equations (1) and (2):

$\angle AOB = 2\angle APO+2\angle BPO\\\angle AOB = 2(\angle APO+\angle BPO)\\\bold{\angle AOB = 2(\angle APB)}$

Hence, proved.