Prove that, ‘The measure of an inscribed angle is half the measure of the arc intercepted by it’.

Answer: \(m\angle ABC = \dfrac{1}{2} m(\text{arc AXC})\) (proved below)

[Figure: circle with centre O; inscribed \(\angle ABC\) intercepts arc AXC; ray BO meets the circle again at D, between A and C on arc AXC.]

Given: \(\angle ABC\) is inscribed in a circle with centre O and intercepts arc AXC. Construction: Draw ray BO meeting the circle at D; draw seg OA and seg OC.

Proof: In \(\Delta OAB\), OA = OB (radii) \(\therefore \angle OAB = \angle OBA\). \(\angle AOD\) is an exterior angle of \(\Delta OAB\), so \(\angle AOD = \angle OAB + \angle OBA = 2\angle OBA\).

\(\angle AOD\) is a central angle, so \(m(\text{arc AD}) = \angle AOD = 2\angle ABD\) \(\therefore \angle ABD = \dfrac{1}{2} m(\text{arc AD})\) ...... (I)

Similarly, from \(\Delta OCB\): \(\angle DBC = \dfrac{1}{2} m(\text{arc DC})\) ...... (II)

Adding (I) and (II): \(\angle ABC = \angle ABD + \angle DBC = \dfrac{1}{2}\left[m(\text{arc AD}) + m(\text{arc DC})\right] = \dfrac{1}{2} m(\text{arc AXC})\). Hence proved.

(When the centre lies on an arm or outside the angle, the same steps with subtraction give the result.)