Weber problem that, at the optimum, the angles between the AD, BD, and CD straight lines must be equal to 360° / 3 = 120°. Torricelli deduced from that conclusion that: 1– if any triangle ABD, whose ∠ADB angle is equal to 120°, generates an ABDE convex quadrilateral inscribed in a circle, the ∠ABE angle of the ABE triangle must be equal to (180° − 120°)= 60°; 2– one way to determine the set of locations of D for which the ∠ADB angle is equal to 120° is to draw an equilateral ABE triangle (because each angle of an equilateral triangle is
Triangle center of all is the line ("t","t","t") which corresponds to the set of all equilateral triangles. The point of concurrence of the perpendicular bisectors of the sides of triangle ABC is the circumcenter. The trilinear coordinates of the circumcenter are Let "f"("a","b","c") = "a"("b" + "c" − "a"). Then so f is a triangle center function. Since the corresponding triangle center has the same trilinears as the circumcenter it follows that the circumcenter is a triangle center. Let A'BC be the equilateral triangle having base BC and vertex A' on the negative side of BC and let AB'C and ABC' be