Upon a request, I am posting theorems related to Poncelet points and antigonal conjugates in Euclidean plane geometry. This is going to be a bunch of theorems, but nothing stops you from seeing it as a bunch of exercises and writing the proofs under my post!

1. Poncelet points

Theorem 1. Let A, B, C, D be four points in the plane. The nine-point circles of triangles ABC, BCD, CDA, DAB concur at one point.

This point is called the Poncelet point of the four points A, B, C, D. Note that there are two cases when the Poncelet point of four points is not defined: In fact, if the quadrilateral ABCD is orthocentric (this means that , and ; equivalently, each of the four points A, B, C, D is the orthocenter of the triangle formed by the other three), then the nine-point circles of triangles ABC, BCD, CDA, DAB coincide, and hence the Poncelet point of the four points A, B, C, D cannot be uniquely defined. Also, if two of the points A, B, C, D coincide, then the Poncelet point of the four points A, B, C, D is not defined.

In the following, a non-orthocentric quadrilateral will mean a quadrilateral that is not orthocentric and has four pairwisely distinct vertices.

Theorem 2. For every non-orthocentric quadrilateral ABCD, there is exactly one rectangular hyperbola passing through the points A, B, C, D, and the center of this hyperbola is the Poncelet point of the points A, B, C, D.

Note that Theorem 2 almost won't be needed anywhere below, by the way - all the results that can be formulated without conics can be proven without them as well, except for Theorem 5 maybe.

Theorem 3. Let ABCD be a non-orthocentric quadrilateral, and let H be the orthocenter of triangle ABC. Then, the Poncelet point of the points A, B, C, D is simultaneously the Poncelet point of the points H, B, C, D, as well as the Poncelet point of the points H, C, D, A, as well as the Poncelet point of the points H, D, A, B, and lies on the nine-point circles of triangles HAD, HBD, HCD.

The pedal circle of a point P with respect to a triangle ABC is defined as the circle passing through the orthogonal projections of the point P on the lines BC, CA, AB.

Theorem 4. The Poncelet point of four points A, B, C, D lies on the pedal circle of the point A with respect to triangle BCD, on the pedal circle of the point B with respect to triangle CDA, on the pedal circle of the point C with respect to triangle DAB, and on the pedal circle of the point D with respect to triangle ABC.

A harder property to which I don't have an elementary proof yet is:

Theorem 5. The Poncelet point of four points A, B, C, D lies on the circle through the three points , and .

2. Antigonal conjugates

Let ABC be a triangle, and D a point which differs from its vertices A, B, C and from its orthocenter. Then, the reflection of the point D in the Poncelet point of the four points A, B, C, D is called the antigonal conjugate of the point D with respect to triangle ABC. That this "antigonal conjugate" transformation is really a conjugation can be seen from the following fact:

Theorem 6. In the plane of a triangle ABC, select a point D that doesn't lie on any of the circumcircles of the triangles BHC, CHA, AHB, ABC, where H is the orthocenter of triangle ABC. Let D' be the antigonal conjugate of the point D with respect to triangle ABC. Then, the antigonal conjugate of the point D' with respect to triangle ABC is the point D again. [If the point D would lie on one of the circumcircles of the triangles BHC, CHA, AHB, ABC, then its antigonal conjugate D', if it exists, would be one of the points A, B, C, H, and thus the antigonal conjugate of this point D' would be not defined.]

Note that it can be hard to prove Theorem 6 just by means of the above definition of antigonal conjugation, but it will easily follow from the next property:

Theorem 7. If D' is the antigonal conjugate of a point D with respect to a triangle ABC, then < BD'C = - < BDC, < CD'A = - < CDA and < AD'B = - < ADB.

Hereby, we use directed angles modulo 180°. Again, we can show:

Theorem 8. The antigonal conjugate D' of a point D with respect to a triangle ABC is simultaneously the antigonal conjugate of the point D with respect to any of the triangles BHC, CHA, AHB, where H is the orthocenter of triangle ABC.

As a consequence of Theorems 7 and 8, we can get:

Theorem 9. If H is the orthocenter of triangle ABC, and D' is the antigonal conjugate of the point D with respect to triangle ABC, then < AD'H = - < ADH, < BD'H = - < BDH, < CD'H = - < CDH.

From Theorem 6, we can conclude:

Theorem 10. If D' is the antigonal conjugate of a point D with respect to a triangle ABC, then the Poncelet point of the four points A, B, C, D is simultaneously the Poncelet point of the points A, B, C, D', and thus it also lies on the nine-point circles of triangles BCD', CD'A and D'AB.

Another crucial property of antigonal conjugates is:

Theorem 11. Let E be the isogonal conjugate of a point D with respect to a triangle ABC, let E' be the inverse of the point E in the circumcircle of triangle ABC. Then, the isogonal conjugate of the point E' with respect to triangle ABC is the antigonal conjugate D' of the point D with respect to triangle ABC.

In other words, if two points are inverse to each other with respect to the circumcircle of triangle ABC, then their isogonal conjugates with respect to triangle ABC are two points which are antigonal conjugates of each other with respect to triangle ABC.

Theorem 12. If A', B', C' are the reflections of a point D in the sidelines BC, CA, AB of a triangle ABC, then the circumcircles of triangles A'BC, B'CA, C'AB and A'B'C' pass through the antigonal conjugate D' of the point D with respect to triangle ABC.

Theorem 13. If A", B", C", D" are the centers of the circumcircles of triangles A'BC, B'CA, C'AB, A'B'C', respectively, then the point D' is the inverse of the point D" in the circumcircle of triangle A"B"C".

3. An application: The Antigray point

As an application of the above theory, we can prove some results about the "Antigray point" of a triangle.

Let's start with the basic facts that don't need any theory:

Theorem 14. Let , , be the excenters of a triangle ABC, and let , , be the reflections of these excenters , , in the lines BC, CA, AB. Then, the circumcircles of triangles , , and the lines , , all have one common point K.

This point K will be called the Antigray point of triangle ABC.

Note that this point K is X(80) in Clark Kimberling's ETC, and I am calling it Antigray point because of the analogy to the definition of the Gray point X(79).

Now, we denote by I the incenter of triangle ABC, and apply the above theory to D = I. We get:

Theorem 15. If I is the incenter of a triangle ABC, then the Poncelet point P of the points A, B, C, I is the Feuerbach point of the triangle ABC, that is, the point of tangency of the nine-point circle and the incircle of triangle ABC. This Feuerbach point P lies on the nine-point circles of triangles IBC, CIA, ABI, HAI, HBI and HCI, on the pedal circle of the point A with respect to triangle IBC, on the pedal circle of the point B with respect to triangle CIA, on the pedal circle of the point C with respect to triangle ABI, and on the circle through the points where the angle bisectors of the angles CAB, ABC, BCA meet the sides BC, CA, AB.

This was quite a plenty of properties of the Feuerbach point, but now we can consider the antigonal conjugate of the point I and we obtain:

Theorem 16. The Antigray point K of triangle ABC is the antigonal conjugate of the incenter I with respect to triangle ABC. This Antigray point K is the reflection of the incenter I in the Feuerbach point P of triangle ABC. The isogonal conjugate of the Antigray point K with respect to triangle ABC is the inverse of the incenter I in the circumcircle of triangle ABC.

Darij