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.

Answer 1

Given : A circle C(O,r) in which 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 when AB is a minor arc and or a semi-circle.
Reflex $\angle$AOB = $\angle$ACB when AB is a major arc.
Construction: Join AB and CO and produce CO to a point D outside the circle.
Proof: There are three cases:
Case-I AB is a minor arc (Figure -a)
Case-II AB is a semicircle (Figure -b)
Case-III AB is a major arc (Figure -c)
Now, in $△$AOC,
OA = OC (Radii of the circle)
$\angle$1 = $\angle$3 (Angles opposite equal sides)
$\angle$5 = $\angle$1 + $\angle$3 (Exterior angle property)
$\Rightarrow$$\angle$5 = $\angle$1 + $\angle$1
$\Rightarrow$$\angle$5 = 2$\angle$1...................(i)
Now, in $△$BOC,
OB = OC (Radii of the circle)
$\angle$2 = $\angle$4 (Angles opposite equal sides)
$\angle$6 = $\angle$2 + $\angle$4 (Exterior angle property)
$\Rightarrow$$\angle$6 = $\angle$2 + $\angle$2
$\Rightarrow$$\angle$6 = 2$\angle$2...................(ii)
Adding equations (i) and (ii), we get:
In figures (a) and (b),
$\angle$5 + $\angle$6 = 2$\angle$1 + 2$\angle$2
$\Rightarrow$$\angle$5 + $\angle$6 = 2($\angle$1 + $\angle$2)
$\Rightarrow$$\angle$AOB = 2$\angle$ACB
In figure (c),
$\angle$5 + $\angle$6 = 2($\angle$1 + $\angle$2)
$\Rightarrow$Reflex $\angle$AOB = 2$\angle$ACB