Question

The angle subtended by an arc at the center is double the angle subtended by it at any point onthe remaining part of the circle prove it.

Answer 1

Theorem: 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

Given:arc ab of a circle with centre o,subtending angles

To prove:

in the given condition, clearly we have three cases:

case (1) AB is a minor arc

case (2) ab is a semicircle

case (3) ab is major arc

Construction:join PO and extend it to a poin Q.

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 = 2OPB ...(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 = 2OPA ...(4)

Adding (2) and (4), we have

QOA + QOB = 2OPA + OPB

∴ AOB = 2(OPA + OPB)

⇒ AOB = 2APB

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

∴ reflex AOB = 2APB

Hence proved

Answer 2

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