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

* *

*give me the proof of this theorem !??*

**Proof: **

We know that, **an exterior angle of a triangle is equal to the sum of the interior opposite angles.**

In ΔOPB,

∠QOB = ∠OPB + ∠OBP ...(1)

OB = OP (**Radius of the circle**)

⇒ ∠OPB = ∠OBP (**In a triangle, equal sides have equal angle opposite to them**)

∴ ∠QOB = ∠OPB + ∠ OPB

⇒ ∠ QOB = 2∠OPB ...(2)

In ΔOPA

∠QOA = ∠ OPA + ∠ OAP ...(3)

OA = OP (**Radius of the circle**)

⇒ ∠OPA = ∠OAP (**In a triangle, equal sides have equal angle opposite to them**)

∴ ∠QOA = ∠OPA + ∠OPA

⇒ ∠QOA = 2∠OPA ...(4)

Adding (2) and (4), we have

∠QOA + ∠QOB = 2∠OPA + ∠OPB

∴ ∠AOB = 2(∠OPA + ∠OPB)

**⇒ ∠AOB = 2∠APB**

For the case 3, where AB is the major arc, ∠AOB is replaced by reflex ∠AOB.

∴ reflex ∠AOB = 2∠APB

Hence proved.

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