Question

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

Answer 1

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.