In the adjoining figure, and are two equal chords of a circle with center Show that lies on the bisector of

In triangle OAB and triangle OAC,

AB = AC [Given]

OB = CO [Radius]

OA = OA [Common]

By side-side-side criterion of congruence

ΔOAB ≅ ΔOAC

∴ ∠OAB = ∠OAC Proved.