Diameter perpendicular to a chord bisects it (and the subtended arcs)

Dependencies: Perpendicular bisector ⇔ equidistant from endpoints, Perpendicular bisector test, Base angles of an isoceles triangle are equal, Central ∠, arc, chord are pairwise equivalent.

Statement

Let $\odot O$ be a circle in the plane and let $AB$ be a chord (not through $O$). Let $NS$ be a diameter of $\odot O$ with $NS \perp AB$, meeting $AB$ at $M$. Then

That is, the diameter through the center perpendicular to the chord bisects the chord, and also bisects each of the two arcs it cuts off (the "upper" and "lower" arcs of $\odot O$ on either side of $AB$).