Question

Prove that the angle subtended by an arc at the centre of a circle is bisected by the radius passing through the mid-point of the arc.

Answer 1

Given: Let a circle have centre O. Let A and B be two points on the circle, and let M be the midpoint of arc AB, so M lies on the circle and arc AM = arc MB.

To Prove: The radius OM bisects the central angle ∠AOB; i.e., ∠AOM = ∠MOB.

Proof [Step-wise]:

1. Since M is the midpoint of arc AB, the two arcs AM and MB are equal.

2. Equal arcs subtend equal chords in the same circle; hence chord AM = chord MB.

3. Consider triangles ΔOAM and ΔOBM.

OA = OB   ...(Radii of the same circle)

OM = OM   ...(Common side)

AM = MB   ...(From step 2)

4. By the SSS congruence criterion, ΔOAM ≅ ΔOBM.

5. Corresponding angles in congruent triangles are equal; therefore ∠AOM = ∠MOB.

OM bisects the central angle ∠AOB, as required.