Diameter ⊥ chord bisects arc

Dependencies: central angle ≡ arc ≡ chord (Central ∠, arc, chord are pairwise equivalent), perpendicular bisector (Perpendicular bisector ⇔ equidistant from endpoints).

Statement

Let $\odot O$ be a given circle, let $AB$ be a chord that is not a diameter, and let $N$ be the midpoint of one of the two arcs subtended by $AB$ — i.e. the point on that arc that is "equidistant in central angle" from the endpoints $A$ and $B$:

Then the line $ON$ joining the arc midpoint $N$ to the centre $O$ — i.e. the diameter through $N$ — is the perpendicular bisector of the chord $AB$: