Simson lines of diametral points

Consider triangle t=ABC and a point D on its circumcircle (c) and its antipodal point E on (c). Project D, E on the sides of (t) and define the Simson lines S(D), S(E) (see Simson.html ). Then the angle of the two Simson lines S(D) and S(E) is a right one and their intersection point is on the Euler circle of t.

That the angle of S(D), S(E) is a right one follows from the discussion in SimsonProperty2.html .
The proof of the other statement follows from the fact that the orthocenter H of t is the similarity center of the Euler circle and the circumcircle of t. To complete the argument draw parallels to S(D), S(E) intersecting at point G on (c). From SteinerLine.html follows that the intersection point F of S(D) and S(E) and the intersections J, I of these lines with HD, HE make a triangle s = IJF, similar to DEF etc. Notice, by the way that diameter IJ of the Euler circle is parallel to the diameter DE of the circumcircle c.

Remark Projecting the points {E,O,D} of the diameter on one side, AB say, we see that the projections of {D,E} are symmetric with respect to the middle of the corresponding side. This could be used to give another proof of the property discussed above.

Simson.html
SimsonGeneral.html
SimsonGeneral2.html
SimsonProperty.html
SimsonProperty2.html
Simson_3Lines.html
SimsonVariantLocus.html
SteinerLine.html
ThreeDiameters.html
ThreeSimson.html
TrianglesCircumscribingParabolas.html
WallaceSimson.html