The proof results by an angle chasing argument indicated above. The two marked angles at J are equal because of the cyclic quadrilateral ABJC.

[1] J coincides with a vertex of the triangle. Then L(J) extends the corresponding altitude of the triangle.

[2] J is the diametral of vertex A. Then L(J) becomes identical with side-line BC of the triangle.

[3] J is the middle of the arc opposite to A. Then L(J) meets line OJ exactly at the middle of side BC and is parallel to the exterior bisector of angle A.

[4] J is the middle of the arc containing A. Then L(J) meets line OJ exactly at the middle of side BC and is parallel to the interior bisector of angle A.

[5] J is the intersection of the altitude from A with the circumcircle. Then L(J) passes through the foot of this altitude and is parallel to the circumcircle-tangent at A.

[6] L(J) and OJ are parallel <==> L(J) passes through the center of the Euler circle of the triangle.

The proofs follow easily from the discussion in SimsonProperty.html , on the inverse problem of finding a point on the circumcircle whose Simson line is parallel to a given direction.

See SteinerLine.html for the discussion on a line parallel to the Simson line and passing through the orthocenter.

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[2] File SimsonProperty2.html discusses the relation of the central angle of two points P,Q on the circumcircle with the angle of the corresponding Simson lines L(P), L(Q).

[3] File SimsonDiametral.html examines the Simson lines of diametral points on the circumcircle.

[4] File SimsonGeneral.html discusses a generalization of these lines in which the point on the circumference is not projected orthogonally on the sides of the triangle but along an arbitrary (fixed) angle phi.

[2] Honsberger, R.

[3] Yiu, P.

Orthopole.html

SimsonDiametral.html

SimsonGeneral.html

SimsonProperty.html

SimsonProperty2.html

SimsonRelated.html

SteinerLine.html

ThreeDiameters.html

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