We are given that:

$\overarc{AB} \cong \overarc{BC} \cong \overarc{CA}$

We need to prove that $\triangle ABC$ is equiangular.

Step-by-Step Proof:

1. Understanding the Given Information:

   • The problem specifies that $\overarc{AB} \cong \overarc{BC} \cong \overarc{CA}$, meaning the arcs subtended by the sides of the triangle $\triangle ABC$ on the circle are congruent.
   • Since the arcs are congruent, the corresponding central angles that subtend these arcs must also be congruent.

2. Apply the Circle Theorem (Equal Arcs Subtend Equal Angles):

   • The angles subtended by arcs at the center of the circle are equal if the arcs themselves are equal.
   • Therefore, the central angles $\angle AOB = \angle BOC = \angle COA$, where $O$ is the center of the circle, are all equal.

3. Inscribed Angles Theorem:

   • The inscribed angle subtended by a chord is half the central angle subtended by the same chord.
   • Since the central angles are equal, the inscribed angles at each vertex of the triangle must also be equal.

   So, $\angle A = \angle B = \angle C$.

4. Conclusion:

   • Since all the angles in $\triangle ABC$ are equal, we conclude that $\triangle ABC$ is equiangular.

Thus, we have proved that $\triangle ABC$ is equiangular.

Would you like more details on any step? Here are some related questions to deepen your understanding:

1. How does the inscribed angle theorem relate to the central angle?
2. Why does congruent arcs imply equal angles in the context of a circle?
3. What is the significance of the circle in making triangle $\triangle ABC$ equilateral?
4. Can an equilateral triangle also be described as equiangular? Why?
5. How would you prove that the sides of $\triangle ABC$ are also equal?

Tip: In any circle, congruent arcs always lead to congruent angles at the circumference. Use this concept when dealing with geometrical proofs in circles!