theorem, this implies that the points are concyclic. So, using the inscribed angle theorem applied to the segment , . Because , the point lies on the line segment . So, the lines are concurrent (they intersect at a single point). Q.E.D. This proof applies only in Case 2, since if , point lies inside the circumcircle of which switches the relative positions of and . However it is easily modified to cover Case 1. Then hence which means is concyclic so . Therefore, lies on . The lines joining the centers of the circles in Fig. 2 are perpendicular to the line segments . For example, the line joining the center of the circle containing and the center of the circle containing , is perpendicular to the segment . So, the lines joining the centers of the circles also intersect at 60° angles. Therefore, the centers of the circles form an equilateral triangle. This is known as Napoleon's Theorem. Location of the Fermat point Traditional geometry Given any Euclidean triangle and an arbitrary point let The aim of this section is to identify a point such that for all If such a point exists then it will be the Fermat point. In what follows will denote the points inside the triangle and will be taken to include its boundary . A key result that will be used is the dogleg rule, which asserts that if a triangle and a polygon have one side in common and the rest of the triangle lies inside the polygon then the triangle has a shorter perimeter than the polygon: If is the common side, extend to cut the polygon at the point . Then the polygon's perimeter is, by the triangle inequality: Let be any point outside . Associate each vertex with its remote zone; that is, the half-plane beyond the (extended) opposite side. These 3 zones cover the entire plane except for itself and clearly lies in either one or two of them. If is in two (say the and zones’ intersection) then setting implies by the dogleg rule. Alternatively if is in only one zone, say the -zone, then where is the intersection of and . So for every point outside there exists a point in such that Case 1. The triangle has an angle ≥ 120°. Without loss of generality, suppose that the angle at is ≥ 120°. Construct the