A diameter ⊥ to a chord bisects the subtended arcs

Dependencies: SAS congruence, central angle ≡ arc ≡ chord (Central ∠, arc, chord are pairwise equivalent).

Statement

Let $\odot O$ be a given circle and let $AB$ be a chord of $\odot O$; let $PN$ be a diameter through the center $O$ with $PN \perp AB$, and let $PN$ meet $AB$ at $M$. Then $PN$ bisects both arcs subtended by the chord:

In other words, "the diameter ⊥ to the chord" splits each of the two arcs between $A$ and $B$ on the circle into two equal halves.