[alogo] 1. Orthopole-preliminary

Start with an interesting exercise.
Consider a triangle ABC and a line (d) through its vertex C (controlled by the point F). Project vertices A and B on this line, to points G, H respectively. Show that the orthogonals GI and HK to the sides BC and AC respectively, intersect at a point L of the (extension of) altitude CJ of the triangle.

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Define points D, E as intersections of the sides of ABC with the lines KH and GI respectively. Show that triangle CDE is (anti) homothetic to ABC, the homothety being centered at C and having a modulus that can be calculated from the given data.
Indeed, triangles IEC and MAC are similar and EC/CA = CI/CM = b*cos(v)*cos(f)/(b*cos(C)) = cos(v)*cos(f)/cos(c) = r. The same ratio is found for DC/CB, proving that r is the modulus of the homothety.

[alogo] 2. Orthopole

Consider a triangle ABC and a line (d). Project vertices A, B and C on this line, to points E, F, G respectively. Show that the orthogonals GI, EH, FJ to sides AB, BC, CA respectively meet at a point D. D is called the orthopole of line d with respect to triangle ABC.

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Consider triangle DEF, formed by the line d and the lines FJ and EH. We have to show that GI passes through D. Notice first that triangle DEF is parallel translated and orthogonally to (d), when line (d) is translated parallel to itself. By such a translation D describes a line parallel to CG at a fixed distance from it. Taking d to pass through C, we reduce the problem to the previous exercise, proving that GD is orthogonal to AB.

[alogo] 3. Relation to Simson-Wallace lines

The orthopole D of line (d) is on the Simson-Wallace line DD' = W(R) which is orthogonal to line (d).

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According to the remark of the previous section moving d parallel to itself makes triangle DEF, involved in the construction of the orthopole, also move parallel to itself. When d obtains the position through G', defined as the second intersection with the circumcircle of the line CG which is orthogonal to (d), then D takes takes the position D' on side AB of the triangle.
In fact, draw from G' a parallel d' to (d) and G'D' orthogonal to AB. Draw also AE', BF' orthogonal to d'. G'F'BD' is cyclic and consequently angle(G'F'D') = angle(G'BD') = angle(G'CA). But last angle is equal to angle(GFJ) since the two angles have their sides pairwise orthogonal. This implies the isometry of triangles GFD and G'F'D and analogously the isometry of triangles EGD and E'G'D'.
By the well-known property (see SimsonProperty.html ) line DD' coincides with the Simson line of point R, which is the other point of intersection of G'D' with the circumcircle.

[alogo] 4. Orthopoles of particular lines

The orthopoles D of lines (d) through the circumcenter lie on the Euler circle of the triangle.

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To see this define the intersection point D'' of the Simson line DD' of R with the altitude HC, H being the orthocenter of ABC. RD''HD0 is a parallelogram (see SimsonProperty.html ) and D' the orthopole of d' coincides with the middle of RH which is a point of the Euler circle of ABC (see Euler.html ).
Remark This property is a special case of the orthopole of a chord of the circumcircle. The Wallace-Simson lines of the extremities of a chord pass both through the orthopole of the chord (see Orthopole2.html ). When the chord is a diameter the corresponding Wallace-Simson lines intersect at right angle on a point of the Euler circle (see SimsonDiametral.html ).

[alogo] 5. References

[1] Gallatly William The modern geometry of the triangle . London, Francis Hodgson 1913, p.49.
[2] Honsberger, R. Episodes in Nineteenth and Twentieth Century Euclidean Geometry. Washington DC, Math. Assoc. Ammer., 1995, pp. 106-110.
[3] Lalesco, T. La Geometrie du Triangle. Paris, Jacques Gabay, 1987, p. 17.

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