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 [tex]\angle APB[/tex] on the circle and [tex]\angle AOB[/tex] on the center of the circle.
To prove:
[tex]\angle AOB = 2 \times \angle APB[/tex]
Proof:
In [tex]\triangle PAO[/tex]: 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, [tex]\angle PAO = \angle OPA[/tex]
Using external angle property, that external angle is equal to sum of two opposite internal angles of a triangle.
[tex]\angle AOQ = \angle PAO + \angle OPA=2 \times \angle APO .... (1)[/tex]
Similarly,
In [tex]\triangle PBO[/tex]: 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, [tex]\angle PBO = \angle OPB[/tex]
Using external angle property, that external angle is equal to sum of two opposite internal angles of a triangle.
[tex]\angle BOQ = \angle PBO + \angle OPB=2 \times \angle BPO .... (2)[/tex]
Now, we can see that:
[tex]\angle AOB = \angle AOQ+\angle BOQ[/tex]
Using equations (1) and (2):
[tex]\angle AOB = 2\angle APO+2\angle BPO\\\angle AOB = 2(\angle APO+\angle BPO)\\\bold{\angle AOB = 2(\angle APB)}[/tex]
Hence, proved.
