Let \( A , B \) be points on the plane, then we can construct the circle from \( A \) to \( B \), \( A \odot B \) and \( B \odot A \), from this we obtain two intersection points and pick one of them as \( C \), and draw a third circle \( C \odot B \).
We know that \( C \in A \odot B \) so \( \text{dist} \left ( A , B \right ) = \text{dist} \left ( A , C \right ) \), additionally \( C \in B \odot A \) so \( \text{dist} \left ( B , C \right ) = \text{dist} \left ( B , A \right ) \), thus we have \( \text{dist} \left ( A , B \right ) = \text{dist} \left ( A , C \right ) = \text{dist} \left ( B , C \right ) \) and so by drawing line segments between \( A , B , C \) we obtain a triangle all with equal side lengths
And thus the triangle \( \overline{A B} , \overline{B C} , \overline{C A} \) is equilateral.