Consider the three disjoint arcs of a circle determined by three points of the circle. We construct a circle around each of the midpoint of every arc which goes the end points of the arc. Prove that the three circles pass through a common point.
1. **Identify the given points and arcs:**
   Let the three points on the circle be \( A \), \( B \), and \( C \). These points determine three disjoint arcs: \( \overset{\frown}{AB} \), \( \overset{\frown}{BC} \), and \( \overset{\frown}{CA} \).

2. **Construct the midpoints of the arcs:**
   Let \( D \), \( E \), and \( F \) be the midpoints of the arcs \( \overset{\frown}{BC} \), \( \overset{\frown}{CA} \), and \( \overset{\frown}{AB} \) respectively.

3. **Construct circles around the midpoints:**
   Construct circles with centers at \( D \), \( E \), and \( F \) that pass through the endpoints of their respective arcs. Specifically:
   - Circle centered at \( D \) passes through \( B \) and \( C \).
   - Circle centered at \( E \) passes through \( C \) and \( A \).
   - Circle centered at \( F \) passes through \( A \) and \( B \).

4. **Prove the concurrency of the circles:**
   We need to show that these three circles pass through a common point. 

   - **Lemma 1:** Let \( I \) be the incenter of \( \triangle ABC \). The midpoint \( D \) of the arc \( \overset{\frown}{BC} \) (not containing \( A \)) satisfies \( DB = DC = DI \). This is because \( D \) is equidistant from \( B \) and \( C \), and the incenter \( I \) is also equidistant from \( B \) and \( C \).

   - **Lemma 2:** Let \( E \) be the midpoint of the arc \( \overset{\frown}{CA} \) (not containing \( B \)). Similarly, \( E \) is equidistant from \( C \) and \( A \), and the incenter \( I \) is also equidistant from \( C \) and \( A \).

   - **Lemma 3:** Let \( F \) be the midpoint of the arc \( \overset{\frown}{AB} \) (not containing \( C \)). Similarly, \( F \) is equidistant from \( A \) and \( B \), and the incenter \( I \) is also equidistant from \( A \) and \( B \).

5. **Conclusion:**
   Since \( D \), \( E \), and \( F \) are the midpoints of the arcs and each of these points is equidistant from the endpoints of their respective arcs and the incenter \( I \), it follows that the circles centered at \( D \), \( E \), and \( F \) all pass through the incenter \( I \).

\[
\boxed{\text{The three circles pass through a common point, which is the incenter of } \triangle ABC.}
\]