# Isogonal locii

Given any triangle ABC, let O the circumcenter and (d) any line through O.
P a moving point on (d). T, U, V the reflections of P through the sides of ABC, and Q the circumcenter of TUV.
Locus of Q when P describes (d).

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We first prove a general property of P and Q, whatever the locus of P (and Q).
BC is the perpendicular bisector of PT, and AB that of PV. Hence B is the circumcenter of PTV. And the perpendicular bisector of TV goes through B and is line BQ.
Similarily, the perpendicular bisector of TU goes through C and is line CQ.
<)PBC = 1/2 PBT = PVT = ABQ, and similarily <)PCA = BCQ
In other words

 Q is the isogonal conjugate of P

Definition :
When P describes a locus (L), the isogonal conjugate Q of P describes a locus (L') just named the isogonal conjugate of (L).

P now describing any line (d), through O or not.

Sorry, your browser is not Java compliant Let Q1 the intersection of BQ and (d). I1, J1 the intersections of angle bisectors of B with (d).
(P,Q1,I1,J1) is an harmonic range, and this defines an involution P ↔ Q1, with fixed points I1 and J1.
Similarily, Q2 intersection of CQ and (d), and angle bisectors of C intersection (d) in I2, J2 define an involution P ↔ Q2.
The composition of these involutions defines a projective homography Q1 → Q2 on line (d) and is projected into an homography between the pencils of lines B* and C* : BQ1 → CQ2.
The Chasles-Steiner theorem then says that the intersection Q of BQ1 and CQ2 describes a conic section, which by the way goes through B and C.
Exchanging the role of A,B,C, the conic section also goes through A.
The applet shows the homographies Q1 → Q2 and BQ1 → CQ2.
(d) is defined by movable points I1, J1 on the angle bisectors of B. Draggable point Q1 is the current point in the homographies.

 The isogonal conjugate of any line (d) is a conic section circumscribed to ABC

#### Points at infinity

Recall the Simson theorem :
Let M,N,P the perpendicular projections of P on the sides of ABC
M,N,P are in line if and only if P is on the circumcircle to ABC.

By a dilation of ratio 2 and center P, points T,U,V are in line in the same condition.
The "circumcircle" of TUV becomes then the Steiner line TUV, and the "circumcenter" Q is the point at infinity in perpendicular direction to TUV.
Depending on the intersections of (d) with the circumcircle, there will be 0, 1 or 2 points at infinity on the conic section, locus of Q.

 The conic section locus of Q is an ellipse, a parabola or an hyperbola depending on the position of (d) with respect to  the circumcircle of ABC.  The asymptotes are perpendicular to the Simson lines of intersection points of (d) with the circumcircle.

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In the applet, line (d) is freely defined by the draggable blue points.
Some tricks allow to draw "clean" hyperbolas and asymptotes :
- Distinguish the cases when (d) intersects the circumcircle or not.
- The intersection points of (d) with the circumcircle are by-passed by separately drawing the locii of Q when P describes
the segment ]P1P2[, the half-line ]P1x[ and the half line ]P2x'[
- The asymptotes are separately constructed by the Pascal construction of a tangent to a conic section defined by 5 points :
points A,B,C and the points at infinity D and E in directions perpendicular to the Simson lines.
The asymptotes being the tangent in D and E.
The construction of Q as isogonal conjugate of the current pointP is not displayed, same as above.
Note : the 4th intersection point of the conic section with the circumcircle is when P is at infinity.

In our case, (d) is a diameter of the circumcircle. The Simson lines of diametrally opposite points are perpendicular and the hyperbola is then rectangular.
When P is in O, Q is at the isogonal point of O, that is the orthocenter H of ABC.
We are not surprised :

 The rectangular hyperbolas circummscribed to triangle ABC make a pencil, through base points A,B,C,H.
In other words, for all lines (d) through O, the hyperbolas all go through H.

At last the Simson lines of the intersection points of (d) with the circumcircle are really the asymptotes themselves.
The hyperbola being defined in a projective way (by the Chasles-Steiner theorem), the most direct proof will be here a projective proof.

Sorry, your browser is not Java compliant Consider the hyperbola H from a given line (d), and let (h) one of its asymptotes.
Let D,E,F the intersection points of the ABC sides with (h).
D',E',F' the intersection points of ABC altitudes with (h).
The projective homographie of (h) on itself defined by (D,E,F) ↔ (D'E'F') is really the Desargues involution φ defined on (h) by the pencil of rectangular hyperbolas with base points A,B,C,H.
This because degenerated hyperbolas of this pencil are made of a side and the related altitude.
The point at infinity of (h) is unchanged in this involution, for it is the tangency point of hyperbola H with its asymptote (h).

Consider then the projection λ through center H of (h) onto the line at infinity.
It transforms points D',E',F' into points at infinity A,B,C, intersections of altitudes with line at infinity.
Being a projection, it holds the point at infinity of (h) invariant, as intersection of the two lines.

Consider now the product α of these two homographies, that is the homography of (h) to the line at infinity defined by (D,E,F) → (A,B,C)
The point at infinity of (h) being unchanged in φ and in λ, it is unchanged by their product α
The homography α holding the intersection point of the two lines is therefore a projection.
In other words the lines connecting homologous points DA, EB, FC are concurrent at the center of this projection.

But DA is the perpendicular from D to BC (parallel to altitude AH)
The perpendiculars in D,E,F to the sides of ABC are then concurrent, which is the reciprocal of the Simson theorem,
that is (h) is exactly the Simson line in direction of (h).
The applet shows the homographies m → m'= φ(m) and the associated conic section (intersecting (h) in m and m'),
m' → m' = λ(m') and finally m → m' = α(m),
The points at infinity are showed as parallel lines m'λ(m') and mα(m), through the same point at infinity m'.

The center ω of the hyperbola (intersection of the perpendicular Simson lines) is on the Euler circle of ABC.
Note : the 4th intersection point S of hyperbola with the circumcircle is the reflection of H through the hyperbola center ω.
For it belongs to hyperbola (reflection of H through ω) and to the circumcircle (dilation of ratio 2 with center H of the Euler circle).

To summarize our case when P describes a line (d) through O :

 The locus of Q is a rectangular hyperbola going through A,B,C, and orthocenter H. Its asymptotes are the Simson lines of intersections points of (d) with the circumcircle.

#### Kiepert

Sorry, your browser is not Java compliant The Kiepert hyperbola is one of our rectangular hyperbolas circumscribed to ABC.
It specifically goes through the centroid G of ABC.
The corresponding point P is then the isogonal conjugate of G, that is the Lemoine point K of ABC, intersection point of the symedians.

The hyperbola is therefore the Kiepert hyperbola when line (d) is line OK, named the Brocard axis of ABC.

 The Kiepert hyperbola is the isogonal conjugate of the Brocard axis OK.

Please note that the origin definition of the Kiepert hyperbola is different :
Construct three isosceles similar triangles, on bases AB,AC,BC and corresponding vertices C', B' A'.
Lines AA',BB',CC' are concurrent in Q and the locus of Q is the Kiepert hyperbola.
Note that the Fermat-Toricelli point is on this hyperbola : when the isosceles triangles are equilateral.

#### Jerabek

The Jerabek hyperbola also belongs to our family of hyperbolas, but goes through O.
It is then the transformed of a line through the isogonal conjugate of O, that is orthocenter H, hence the Euler line OH.
The prominent role of the Euler line in triangle ABC gives a similar importance to its isogonal conjugate, that is the Jerabek hyperbola. See the link below.
We specifically note here that G being on the Euler line, the conjugate of G, that is the Lemoine point K, is on the Jerabek hyperbola.

#### Feuerbach

The Feuerbach hyperbola goes through incenter I which is its own isogonal conjugate. This hyperbola is then the isogonal conjugate of line OI.
Its center is at the Feuerbach point, contact point of Euler circle and incircle.

A Web page giving many interresting properties of the Jerabek, Kiepert and Feuerbach hyperbolas, as well as other related conic sections, is here. Impressive !