Question

Please prove that angle suntended by an arc at centre is double the angle subtended by it at 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. this is your answer plz mark it as brainlist answer