In , let , , be the feet of altitudes from , , respectively. Let be the orthocenter of . Let , , be the projections of points , , , on , , respectively. Assume that the circumcircles of and intersect at two points and , prove that .

We solve it only for acute because the other cases are similar. Define the feet from to respectively. Notice that points are useless and we just want to show that is collinear with the circumcenters of and Thus we call the medial triangle of

First notice by Reim's on and Then Reim's again on gives is cyclic. Now its center lies on the perpendicular bisectors of and Notice that the perpendicular bisector of bisects and thus contains and furthermore that the internal angle bisector of is perpendicular to since and Thus the center of must be the incenter of and by symmetry we get is cyclic with center

Let be the circumcenter of We wish to show that are collinear. In fact we claim that is the reflection of over Now is by definition the Bevan point of so it is the reflection of the incenter of over its circumcenter However since are the incenter and orthocenter of the medial triangle of the orthic triangle, we get with directed lengths, by homothety. Thus the reflection of over is the same as the reflection of over which is done.