Prove that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. Using this theorem calculate angle AEB i n fig if AB is a diameter and chord CD is equal to radius of circle AC and BD are produced to intersect at E.

The question asked by you which is associated with the theorem is not complete.

But the proof for the theorem is as follows-

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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