Solution:

Two chords of a circle through equally spaced points have equal length exactly when they skip the same number of vertices, so all $12$ segments join pairs of vertices exactly $k$ apart for one common $k \in \{1, \ldots, 12\}.$ For fixed $k,$ form the graph on the $24$ vertices joining each $i$ to $i \pm k \pmod{24}:$ we need a perfect matching in this graph. For $k \lt 12$ the graph is a disjoint union of $\gcd(24, k)$ cycles of length $24/\gcd(24, k),$ while for $k = 12$ it is $12$ disjoint diameters.

A cycle of even length has exactly $2$ perfect matchings (alternate edges), and a cycle of odd length has none. So each $k \lt 12$ with even cycle length contributes $2^{\gcd(24, k)}:$ $k = 1, 5, 7, 11$ give $2$ each; $k = 2, 10$ give $4$ each; $k = 3, 9$ give $8$ each; $k = 4$ gives $16;$ $k = 6$ gives $64.$ For $k = 8$ the cycles have odd length $3,$ giving $0.$ For $k = 12$ the matching is forced: $1$ way.

The total is $4 \cdot 2 + 2 \cdot 4 + 2 \cdot 8 + 16 + 64 + 0 + 1 = 113.$