AB is a chord of a circle having centre O. If ∠AOB = 60°, then prove that the chord AB is of radius length.
Here, OA = OB = r [radii of same circle]
⇒ ∠A = ∠B —– (i)
∠O + ∠A + ∠B = 180°
⇒ 60° + ∠A +∠A = 180° [using eq. (i)]
⇒ ∠A = 60°
Thus ∠O = ∠A = ∠B = 60°
⇒ ΔOAB is an equilateral triangle .
⇒ AB = OA = OB = r Hence proved.