In triangle with incenter and circumcenter , let be the points of tangency of its circumcircle with its -mixtilinear circles, respectively. Let be the circle through that is tangent to at , and define similarly. Prove that have a common point other than .

Proof:

Lemma 1: If , are the tangency points of the -mixtilinear incircle with respectively then is the midpoint of .

Proof: This is a well-known result, and moreover can be proved immediately with a limiting case of Sawayama's Theorem, but I shall provide a more elementary proof. Let be the midpoints of arcs respectively. Then by Archimedes' Lemma we have that and are collinear. Then by Pascal's Theorem on hexagon we have that are collinear. But since and this immediately implies the desired result.

Lemma 2: If denotes the midpoint of arc then are collinear

Proof: Consider the homothety centered at that takes the -mixtilinear incircle to the circumcircle of triangle . This takes to and to so by Lemma 1 it takes to , the midpoint of . So it suffices to show that are collinear. Some quick angle-chasing yields the fact that is a parallelogram which implies the desired result.

Now, taking the inversion centered at with radius , we have that is mapped to the line parallel to passing through the reflection of over . We obtain two similar lines for and . Now, reflect all three lines about .

So, letting be the line through parallel to and defining similarly it suffices to show that these three lines concur at the circumcenter of the excentral triangle of triangle . Let be the -excenters of triangle respectively. It is well-known that is the midpoint of and that so is the perpendicular bisector of . This immediately implies the desired result, and so we are done.