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.

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 $\angle$AOB at the centre and $\angle$APB at a point P on the remaining part of the cirlce.

To prove: $\angle$AOB = 2$\angle$APB

In the given condition, clearly we have three cases:
Case I: Arc AB is a minor arc
Case II: Arc AB is a semicircle
Case III: Arc AB is a major arc

Construction: Join PO and produce it to a point Q.


Proof:

In $∆$AOP, $\angle$AOQ is an exterior angle
As, an exterior angle is equal to the sum of the interior opposite angles.
So, $\angle$AOQ = $\angle$APO + $\angle$PAO .....(i)

Also, AO = PO (Radii)
So, $\angle$APO = $\angle$PAO (Angles opposite to equal sides are equal) .....(ii)

From (i) and (ii), we get
$\angle$AOQ = $\angle$APO + $\angle$APO
$\Rightarrow$$\angle$AOQ = 2$\angle$APO .....(iii)

Similalry,
In $∆$BOP, $\angle$BOQ is an exterior angle
As, an exterior angle is equal to the sum of the interior opposite angles.
So, $\angle$BOQ = $\angle$BPO + $\angle$PBO .....(iv)

Also, BO = PO (Radii)
So, $\angle$BPO = $\angle$PBO (Angles opposite to equal sides are equal) .....(v)

From (iv) and (v), we get
$\angle$BOQ = $\angle$BPO + $\angle$BPO
$\Rightarrow$$\angle$BOQ = 2$\angle$BPO .....(vi)

Adding (iii) and (vi), we get
$\angle$AOQ + $\angle$BOQ = 2$\angle$APO + 2$\angle$BPO
$\Rightarrow$$\angle$AOB = 2($\angle$APO + $\angle$BPO)
So, $\angle$AOB = 2$\angle$APB

Note: In Case III, where AB is a major arc, $\angle$AOB is replaced by reflex $\angle$AOB.
So, reflex $\angle$AOB = 2$\angle$APB