Given a pair of concentric circles, chords  $AB,BC,CD,\dots$ of the outer circle are drawn such that they all touch the inner circle. If $\angle ABC = 75^{\circ}$, how many chords can be drawn before returning to the starting point ?
[img]https://i.imgur.com/Cg37vwa.png[/img]
1. **Identify the given angle and its implications:**
   Given that $\angle ABC = 75^\circ$, and since all chords touch the inner circle, each subsequent angle $\angle BCD, \angle CDE, \ldots$ will also be $75^\circ$.

2. **Determine the central angle:**
   The central angle formed by connecting the radii to consecutive tangency points is calculated as:
   \[
   180^\circ - 75^\circ = 105^\circ
   \]
   This is because the angle between the chord and the tangent at the point of tangency is supplementary to the angle subtended by the chord at the center of the circle.

3. **Calculate the number of chords:**
   To find the number of such chords that can be drawn before returning to the starting point, we need to determine how many times the central angle $105^\circ$ fits into a full circle $360^\circ$. This is given by:
   \[
   \frac{360^\circ}{\gcd(360^\circ, 105^\circ)}
   \]
   First, we find the greatest common divisor (gcd) of $360$ and $105$:
   \[
   \gcd(360, 105) = 15
   \]
   Therefore, the number of chords is:
   \[
   \frac{360^\circ}{15^\circ} = 24
   \]

Conclusion:
\[
\boxed{24}
\]