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 d(P_0=|AD|. Moreover, if P\ne P_0 then either or won't lie on which means d(P_0)=|AD| Now allow to range outside . From above a point P' \in \Omega exists such that d(P') and as d(P_0)\leq d(P') it follows that d(P_0) for all outside . That means is the Fermat point of . In other words, the Fermat point is coincident with the first isogonic center.
