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].

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'H

(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.

OrthicAndPedal.html

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WallaceSimson.html

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