[alogo] Steiner line

Consider triangle ABC and a point D on its circumcircle. The projections E, F, G of D on the sides of the triangle are on a line L(D), the Wallace-Simson line of D (look at Simson.html ). Extend segments DE, DF, DG to their doubles. The end points of the new segments define a parallel S(D) to L(D). S(D) passes through the orthocenter of the triangle.

[0_0] [0_1] [0_2] [0_3]
[1_0] [1_1] [1_2] [1_3]

In fact, it is plain that this line S(D) = [HN] is parallel to the Simson line L(D). Draw the altitude BQ to side AC and build the parallelogram HNDP. The equality of angles at R, C and F shows that BRDP is also a parallelogram. Hence HNDQ is an isosceles trapezium and MF, being orthogonal to its basis DN at its middle, is medial line of HQ. This identifies H as the orthocenter on BM. The symmetric S of D with respect to the Simson line is also on line [HN].

Corollary-1 If follows that the middle K (not drawn) of segment DH is on L(D).

Corollary-2 The reflected of the Steiner line S(D) on the sides of the triangle are lines passing through D.

In fact, extend DE and SH until they intersect at D'. By the parallelity at double distance to L(D) E is the middle of DD'. Hence triangle DD'HC, is isosceles, HC being the intersection point of S(D) with AB. This proves the claim for the reflected of S(D) on AB. Analogously are proved the other concurences at D.

Remark-1 The discussion shows that the Steiner line S(D) of D could be defined as the line carying the reflexions of D on the sides of the triangle.

Remark-2 The discussion leads also to two ways of finding a point D on the circumference, for which the line L(D) has a given direction:
(1) Draw BR parallel to the given direction, to find R, and draw an orthogonal RD to AC (opposite to B), to find D.
(2) Reflect S(D) on the sides of ABC and get the concurence point of the reflected lines.

See Also



Lalesco, T. La Geometrie du Triangle. Paris, Jacques Gabay, 1987, p. 8

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