Theorem

Let triangles be erected externally on the sides of a ABC so that the sum of the "remote" angles P,Q, and R is 180^o. 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 ACQ and BCP. BFC + P = 180^o. Also, AFC + Q = 180^o. Combining these with P + Q + R = 180^o immediately yields AFB + R = 180^o. So that F also lies on the circumcircle of ABR.

This simple fact has the following

Corollary 1

If the vertices A,B,C of ABC lie, respectively, on sides QR, RP, and PQ of PQR, 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 ABC, 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 O_P, O_Q, and O_R be the circumcenters of BCP, ACQ, and ABR, respectively. Consider O_PO_QO_R. Since its sides are perpendicular to the common chords of the three circles, O_P = P, O_Q = Q, and O_R = 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.

Remark