Prove that the angle subtended by an arc at the centre is double the angle subtended by it at any - Maths -

Question

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

Ayush Jha · Accepted Answer

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 Î”OPB,

âˆ QOB = âˆ OPB +
âˆ OBP (1)

OB = OP (Radius of the circle)

â‡’ âˆ OPB =
âˆ OBP (In a triangle, equal sides have
equal angle opposite to them)

âˆ´ âˆ QOB =
âˆ OPB + âˆ  OPB

â‡’ âˆ  QOB =
2âˆ OPB (2)

In Î”OPA

âˆ QOA = âˆ  OPA +
âˆ  OAP (3)

OA = OP (Radius of the circle)

â‡’ âˆ OPA =
âˆ OAP (In a triangle, equal sides have
equal angle opposite to them)

âˆ´ âˆ QOA =
âˆ OPA + âˆ OPA

â‡’ âˆ QOA =
2âˆ OPA (4)

Adding (2) and (4), we have

âˆ QOA + âˆ QOB =
2âˆ OPA + âˆ OPB

âˆ´ âˆ AOB =
2(âˆ OPA + âˆ OPB)

â‡’ âˆ AOB =
2âˆ APB

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

âˆ´ reflex âˆ AOB =
2âˆ APB

Hence proved

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.