Let $d$ be the difference between $\angle B$ and $\angle A,$ which is also the difference between $\angle C$ and $\angle B.$ Then $\angle A = \angle B - d$ and $\angle C = \angle B + d.$ The angles of a triangle always add up to $180^\circ,$ so
\[(\angle B - d) + \angle B + (\angle B + d) = 180^\circ.\]Then $3 \angle B = 180^\circ,$ so $\angle B = 60^\circ.$
Since $\angle A = 23^\circ,$ $d = \angle B - \angle A = 60^\circ - 23^\circ = 37^\circ.$ Therefore, $\angle C = \angle B + d = 60^\circ + 37^\circ = \boxed{97^\circ}.$
