equilateral triangle and for any point in (except itself) construct so that the triangle is equilateral and has the orientation shown. Then the triangle is a 60° rotation of the triangle about so these two triangles are congruent and it follows that which is simply the length of the path . As is constrained to lie within , by the dogleg rule the length of this path exceeds Therefore, for all Now allow to range outside . From above a point exists such that and as it follows that for all outside . Thus for all which means that is the Fermat point of . In other words, the Fermat point lies at the obtuse-angled vertex. Case 2. The triangle has no angle ≥ 120°. Construct the equilateral triangle , let be any point inside , and construct the equilateral triangle . Then is a 60° rotation of about so which is simply the length of the path . Let be the point where and intersect. This point is commonly called the first isogonic center. Carry out the same exercise with as you did with , and find the point . By the angular restriction lies inside . Moreover, is a 60° rotation of about , so must lie somewhere on . Since it follows that lies between and which means is a straight line so Moreover, if then either or won't lie on which means Now allow to range outside . From above a point exists such that and as it follows that for all outside . That means is the Fermat point of . In other words, the Fermat point is coincident with the first isogonic center. Vector analysis Let be any five points in a plane. Denote the vectors by respectively, and let be the unit vectors from along . Adding gives If meet at at angles of 120° then , so for all . In other words, and hence is the Fermat point of . This argument fails when the triangle has an angle because there is no point where meet at angles of 120°. Nevertheless, it is easily fixed by redefining and placing at so that . Note that because the angle between the unit vectors is which exceeds 120°. Since the third inequality still holds, the other two inequalities are unchanged. The proof now continues as above (adding the three inequalities and using )