Consider a triangle t = ABC and a fixed point P, different from the orthocenter H of the triangle. Let X be arbitrary (can take it on the circumcircle of ABC) and reflect line PX on the sides of ABC. In the following facts we consider X varying on the circumcircle of ABC: [1] The reflected lines of PX pass through the reflected points {PA,PB,PC} of P on the respective sides of ABC. Triangle tP = PAPBPC is homothetic (w.r. to P and in ratio 2) to the pedal triangle of P. [2] Triangle t* = A*B*C* formed by these reflected lines is similar to the orthic triangle A'B'C' of ABC. [3] Every point rigidly attached to A*B*C* (i.e. having fixed barycentric coordinates with respect to it) describes a circle. All these circles pass through a point T(P) lying on the circumcircle of ABC. [4] In particular, the circle described by the incenter I of A*B*C* is the circumcircle of ABC. [5] In particular, each vertex of A*B*C* (B* say) describes a circle passing through T(P), but also through one vertex of the triangle (B in this case), two of the reflected points of P on the sides (PA, PC in this case) and two intersection points of the sides with the orthogonals to them from P (BA, BC in this case). [6] Each line rigidly attached to A*B*C* passes through a fixed point. [7] In particular, the bisectors {A*I,B*I,C*I} of triangle t* pass through the respective vertices {A,B,C} of t. As a result {A,A*,I}, {B,B*,I}, {C,C*,I} are aligned and t, t* are perspective. [8] The perspectivity axis of t, t* is line PX. [9] Point T(P) is the point having line HP as Steiner line with respect to ABC. For X on this line the corresponding triangle t* degenerates to point T(P). [10] The maximal triangle A*B*C* (in perimeter/area) occurs when PX is orthogonal to PH. [11] Triangle tO = OAOBOC, having vertices the centers of the three circles described by {A*,B*,C*} respectively, is similar to the orthic of ABC. In fact it is homothetic to the maximal of A*B*C* with respect to the homothety having center at T(P) and ratio 1/2. The circumcenter of ABC coincides with the incenter of tO.
From the respective definitions, follow [1] and [8], i.e. that t and t* are axis perspective with respect to line PX. An easy angle-chasing shows that quadrangle BPAB*PC is cyclic, having its angle at B equal to 2B. Corresponding properties for the angles A* and C* prove [2]. First part of [3] as well as [6] follow from general properties of triangles varying so that they remain similar to theirselves and having sides passing through three fixed points. An excelent discussion on these properties can one find in the book by Yaglom referred below. The second part of [3] and [9], about the circles passing through T(P) on the circumcircle, follows from the fact discussed in SteinerReflected.html , according to which, for line PX passing through the orthocenter H all three points {A*,B*,C*} coincide with one point T(P) of the circumcircle, which is then the point for which PX is the corresponding Steiner line. From its proper definition t* is line perspective to t with perspectivity axis line PX. Thus, lines {AA*,BB*,CC*} concur at a point I. Consider triangle A1A2A*, whose sides are PX and its reflexions on two sides of ABC. Because of the reflexions lines AA1=AC and AA2=AB are bisectors of the angles at A1, A2 respectively. Hence A is incenter of triangle A1A2A* and consequently AA* bisects angle A* of t*. This shows [4], [7] and [8]. The fact that I lies on the circumcircle follows by measuring the corresponding angle AIC. [10] By the discussion in MaximalSegment.html follows that A*C* becomes maximal when P*, C* obtain the positions of diametrals of T(P) with respect to the corresponding circles that carry these points. Then their line L passing through PB is reflected on AC to obtain the direction L' for which this maximum occurs. L' is orthogonal to PH because L is orthogonal to line PBT(P). By the reflection on AC point PB maps to P and T(P) maps to a point P' on its Steiner line which is PH. The similarity of the triangle of centers with the orthic follows from the remark on the reference. By this remark namely triangle tO coincides with the homothetic of the maximal A*B*C* with respect to T(P) at ratio 1/2. Since the incenter of the maximal is on the circumcircle its homothetic by this homothety must coincide with the circumcenter.