Question

(Prove) 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.​

Answer 1

Hope you could understand.

If you have any query, feel free to ask.

Answer 2

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.

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 * \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 * \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 * \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.