Question

Prove that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle. [4 MARKS]

Answer 1

Concept: 2 Marks
Application: 2 Marks

Given : A circle with centre O and an arc AB subtends $\angle AOB$ at the centre and $\angle ACB$ at any point C
on the remaining part of the circle.

To prove : $\angle AOB=2\angle ACB$.

Construction :Join CO and produce it to some point D.

In $\Delta AOC$, we have

OA = OC [Radii of the same circle.]

$\angle OAC=\angle OCA$ ..... (1) [Angles opposite to equal sides of a $\Delta$ are equal.]

$\angle AOD=\angle OAC+\angle OCA$ [Exterior angle property]

$\angle AOD=2\angle OCA$ ......(2) (from (1))

Similarly, In figure (i), we have

$\angle BOD=2\angle OCB$ ...(3)

Adding (2) and (3)

$\angle AOD+\angle BOD=2\angle OCA+2\angle OCB$

$2(\angle OCA+\angle OCB)=2\angle ACB$

$\therefore \angle AOB=2\angle ACB$

In Figure (ii), we have :

$\angle BOD-\angle AOD=2\angle OCB-2\angle OCA$ Subtracting (2) from (3)

$=2(\angle OCB-\angle OCA)$

$=2\angle ACB$

$\angle AOB=2\angle ACB$.

Hence, $\angle AOB=2\angle ACB$.

In Figure (iii), we have :

$\angle AOD+\angle BOD=2\angle OCA+2\angle OCB$ Adding (2) and (3)

$=2(\angle OCA+\angle OCB)$

$=2\angle ACB$

Reflex $\angle AOB=2\angle ACB$.