On each side of a triangle, erect an equilateral triangle, lying exterior to the original triangle. Then the segments connecting the circumcenters of the three equilateral triangles themselves form an equilateral triangle.
Theorem
Let triangles be erected externally on the sides of ΔABC so that the sum of the "remote" angles P,Q, and R is 180°. Then the circumcircles of the three triangles ABR, BCP, and ACQ have a common point.
Proof
Let F be the second point of intersection of the circumcircles of ΔACQ and ΔBCP. ∠BFC + ∠P = 180°. Also, ∠AFC + ∠Q = 180°. Combining these with ∠P + ∠Q + ∠R = 180° immediately yields ∠AFB + ∠R = 180°. So that F also lies on the circumcircle of ΔABR.
This simple fact has the following
Corollary 1
If the vertices A,B,C of ΔABC lie, respectively, on sides QR, RP, and PQ of ΔPQR, then the three circles PCB, CQA, and BAR have a common point.
Corollary 2
If similar triangles PCB, CQA, and BAR are erected externally on the sides of ΔABC, then the circumcircles of these three triangles have a common point. (Please note that the triangles are named in such a manner as to make it obvious that vertices P,Q, and R do not correspond to each other in the given similar triangles.)
Under the conditions of Corollary 2, let OP, OQ, and OR be the circumcenters of ΔBCP, ΔACQ, and ΔABR, respectively. Consider ΔOPOQOR. Since its sides are perpendicular to the common chords of the three circles, ∠OP = ∠P, ∠OQ = ∠Q, and ∠OR = ∠R. Hence we obtain a proof for a generalization of Napoleon's theorem:
Corollary 3
If similar triangles PCB, CQA, and BAR are erected externally on the sides of a triangle ABC, their circumcenters form a triangle similar to the given three triangles.
