For this, consider the shortest side of the triangle, BC say, and draw the tangents to its circumcircle at {B,C} intersecting at A'. Draw also the symmetric A* of A with respect to the middle M of BC and line A'A*, intersecting the sides at {B',C'} correspondingly. The Steiner point S coincides with the intersection point of lines BC' and CB'.

The result is a special case of the method of finding intersection-points of two conics circumscribing ABC, discussed in TriangleConicsIntersections.html . All lines B'C' pivoting at A' define through the intersection-points of {BC',CB'} points of the circumcircle. Analogously all lines B'C' pivoting at A* define through the intersection-points of {BC',CB'} points on the Steiner outer-ellipse. Thus the line through A'A* defines a point lying simultaneously on the two curves.

Maclaurin.html

Steiner_Ellipse.html

TriangleConics.html

TriangleConicsIntersections.html

Produced with EucliDraw© |